My Approach is this- I used an arraylist. I checked all the nodes of the Binary Tree, to find those whose left and right node are null, which signifies that they are a leaf, then I found out their levels and added them to an arraylist.
After this I used the Arraylist in the function boolean check to check whether all the elements of the array list are same of not, if they are i return true (all leaves are at the same level) otherwise I return false.
class Solution {
boolean check(Node root) {
int c = 0;
ArrayList<Integer> a = new ArrayList<>();
for (int x : a) {
if (x != (a.get(0)))
return false;
}
return true;
}
public void che(Node root, int level, ArrayList<Integer> a) {
if (root == null) return;
if (root.left == null && root.right == null) {
a.add(level);
}
che(root.right, level + 1, a);
che(root.left, level + 1, a);
}
}
This is the link for the Question
The che function is never called.
It is however not necessary to collect data in an array list. Instead make the recursive function return the height of the subtree it is called on. In the same function compare the height that is returned for the left and right subtree. If they are different, return a special value to indicate failure (like -2), otherwise return that common height plus one.
This allows the function to abort the search as soon as a height difference is found, avoiding the unnecessary traversal of the rest of the tree.
Here is how that would look:
class Solution
{
boolean check(Node root) {
return height(root) > -2;
}
private int height(Node root) {
if (root == null) return -1;
int left = height(root.left);
if (left == -2) return -2;
int right = height(root.right);
if (left == -1 || right == -1 || left == right) {
return 1 + Math.max(left, right);
}
return -2;
}
}
I was going through a sum on geeksforgeeks.com for adding two linked lists. And I am confused in the answer provided.
Node addTwoLists(Node first, Node second) {
Node res = null; // res is head node of the resultant list
Node prev = null;
Node temp = null;
int carry = 0, sum;
while (first != null || second != null) //while both lists exist
{
// Calculate value of next digit in resultant list.
// The next digit is sum of following things
// (i) Carry
// (ii) Next digit of first list (if there is a next digit)
// (ii) Next digit of second list (if there is a next digit)
sum = carry + (first != null ? first.data : 0)
+ (second != null ? second.data : 0);
// update carry for next calulation
carry = (sum >= 10) ? 1 : 0;
// update sum if it is greater than 10
sum = sum % 10;
// Create a new node with sum as data
temp = new Node(sum);
// if this is the first node then set it as head of
// the resultant list
if (res == null) {
res = temp;
} else // If this is not the first node then connect it to the rest.
{
prev.next = temp;
}
// Set prev for next insertion
prev = temp;
// Move first and second pointers to next nodes
if (first != null) {
first = first.next;
}
if (second != null) {
second = second.next;
}
}
if (carry > 0) {
temp.next = new Node(carry);
}
// return head of the resultant list
return res;
}
I understand that we have created three Nodes res, prev and temp and I don't understand how each one of them are getting updated simultaneously.
if (res == null) {
res = temp;
} else // If this is not the first node then connect it to the rest.
{
prev.next = temp;
}
Like here, how the 2nd element would be added in res when we are adding it in prev.next.
And below :
if (carry > 0) {
temp.next = new Node(carry);
}
If we are adding the last element in temp, how will it reflect in res?
The code seems to be working but I am having a hard time understanding this concept.
Please help. Thanks in advance.
res, prev and temp are references to a Node. When you write something like res = tmp; it means that now res and tmp refer to the same Node instance, and any changes made to this instance using, for example, res reference will remain when you use tmp reference.
here is the same code but with additional comments to try to help explain what is happening. This code appears to be taking two head nodes from link lists where each node contain a single digit, then adding the digits together to create a new list of single digits (e.g. 1->2->3 and 5->6->7->8 with = 1+5->2+6->3+7->0+8. Note that the third set results in a carry over so the final values in the returned list would be 6->8->0->9).
Thus, this code is not adding to two link lists together, it is adding the data contained
in each of the nodes of the two lists and putting the result into a third list.
My additional comments in the code are prefixed with WCK -
Node addTwoLists(Node first, Node second) {
Node res = null; // res is head node of the resultant list
Node prev = null;
Node temp = null;
int carry = 0, sum;
while (first != null || second != null) //while both lists exist WCK actually this is While at least one of the lists exists, I believe the comment is wrong and the code is right
{
// Calculate value of next digit in resultant list.
// The next digit is sum of following things
// (i) Carry
// (ii) Next digit of first list (if there is a next digit)
// (ii) Next digit of second list (if there is a next digit)
sum = carry + (first != null ? first.data : 0)
+ (second != null ? second.data : 0);
// update carry for next calulation
carry = (sum >= 10) ? 1 : 0; // WCK - assumes sum <20; is only carrying at most 1
// update sum if it is greater than 10
sum = sum % 10;
// Create a new node with sum as data
temp = new Node(sum);
// if this is the first node then set it as head of
// the resultant list
// WCK - the first time through the loop, res will be null
// WCK - each subsequent time through the loop, the else is invoked to link the new node to the one from the previous iteration thus linking them
if (res == null) {
res = temp; //WCK - as #ardenit says, both res and temp point to the same node at this point
} else // If this is not the first node then connect it to the rest.
{
prev.next = temp;
}
// Set prev for next insertion
// WCK - now they are starting to set up for the next iteration.
// WCK - the first time through, res, prev and temp all point to the same node after this statement is executed.
prev = temp;
// Move first and second pointers to next nodes
if (first != null) {
first = first.next;
}
if (second != null) {
second = second.next;
}
// WCK - the first time through the loop, both of the lists passed in
// have been advanced to the next node in their respective lists.
// res and prev will point to the temp. As the next iteration of the loop
// starts, temp will be pointed to a new node while res and prev will
// still point to the same node created in the first iteration. As the
// second iteration executes, prev is set to point to the new temp
// where you see (prev.next = temp) and then prev is advanced to the
// this new node as well. res will not be changed again (it is only set
// once in the first iteration)
}
// WCK - now that the loop is finished, there may be a final carry value
if (carry > 0) {
temp.next = new Node(carry);
}
// return head of the resultant list
return res;
}
So in a given binary tree (node binary search tree), initially I want to find the number of nodes between two nodes "p" and "q". I first find the lowest common ancestor between these two nodes, say, "ancestor". Then I calculate the number of nodes between "ancestor" and "p" and number of nodes between "ancestor" and "q" separately and add them at last.
I tried recursive way to get number of nodes between "ancestor" and "p" or "q" but failed. Not a fan of recursive.
public static int NodeToNodePath(BinaryTree root, BinaryTree node, int length){
if(root == null && node == null)
return 0;
if(root == null || node == null)
return 0;
if(root.rootElement == node.rootElement){
return length;
}
int sum = NodeToNodePath(root.left, node, length + 1);
if(sum != 0)
return sum;
sum = NodeToNodePath(root.right, node, sum);
return sum;
}
But in this way, the result from root to left mode is correct but can't find node on the other size.
Any help?
Thanks!
I figure out how to solve the problem. By recursion. This may not apply to my origin problem
"find node number between two given nodes" but it does return number of nodes between root and a given node.
Code is posted below.
public static int NodeToNodePath(BinaryTree root,
BinaryTree node, int length) {
if(root == null)
return 0;
if(root.rootElement == node.rootElement){
length += 1;
return length;
}
int left = NodeToNodePath(root.left, node, length);
if(left != 0){
return left + 1;
}
int right = NodeToNodePath(root.right, node, length);
if(right != 0){
return right + 1;
}
return 0;
}
Also, I found a post on GeekForGeeks talking specifically about my origin problem, it's called "Find distance between two given keys of a Binary Tree" the address is:
http://www.geeksforgeeks.org/find-distance-two-given-nodes/
Thanks everyone!
In fact this is a interview question asked a few days ago.
The interviewer wants me to express the difference between ArrayList and LinkedList, and asked to optimize the insertion operation on ArrayList, in other words, to re-implement add(int index, E element) and of course the complexity of get(int index) operation can be sacrificed.
My answer was to separate the array into k sub-arrays and update a counting array representing the number of elements already in the corresponding sub-array. And the memory of every sub-array is allocated dynamically with an expected initial size. When I need to insert a data into the ArrayList, I can locate a sub-array first, and do the operation within a small array.
And if insertions are not too frequent or the indexes are uniform distributed, the time complexity of inserting can be O(log(k) + n/k + k) in average, where log(k) means we should locate the sub-array first with binary searching on the counting array's sum array, n/k is for data movement or even memory re-allocation, and k stands for the updating of the sum array.
I'm sure there are better solutions. I do need some suggestions, thanks!
One of the solutions could be:
add(int index, E element) always add element to the end of array (you have to also store the index where this element should be added) - complexity O(1)
get(int index) has to restore correct order of array (if some elements were added after the last invocation) - knowing the positions in which each element should be, you can restore correct order in O(n)
You can implement it in a balanced binary tree, so that both add() and get() cost O(logn)
An example implementation will look like (hand-crafted here, will not compile, corner cases not covered):
class Node {
int subTreeSize;
Node left,right;
Element e;
// all i 0-indexed
Node get(int i) {
if (i >= subTreeSize) {
return null;
}
if (left != null) {
if(left.subTreeSize > i) {
return left.get(i);
} else {
i -= left.subTreeSize;
}
}
if (i == 0) {
return this;
}
return right.get(i-1);
}
// add e to the last of the subtree
void append(Element e) {
if(right == null){
right = new Node(e);
} else {
right.append(e);
right = right.balance();
}
subTreeSize += 1;
}
// add e to i-th position
void add(int i, Element e) {
if (left != null) {
if(left.subTreeSize > i) {
add(i,left);
left=left.balance();
} else {
i -= left.subTreeSize;
}
}
if (i == 0) {
if (left == null){
left = new Node(e);
} else {
left.append(e);
left = left.balance();
}
} else {
if (right == null) {
// also in this case i == 1
right = new Node(e);
} else {
right.add(i-1, e);
right = right.balance();
}
}
subTreeSize += 1;
}
// the common balance operation used in balance tree like AVL or RB
// usually just left or right rotation
Node balance() {
...
}
}
public class Tree {
Node root;
public Element get(int i) {
return root.get(i).e;
}
public void add(int i, Element e) {
if (root == null) {
root = new Node(e);
} else {
root.add(i,e);
root = root.balance();
}
}
}
A variant of an order statistic tree would allow you to add and get by index in O(log n).
The basic idea is as follows:
Have each node store the size of the subtree rooted at that node.
The index of a node will correspond to its position in the in-order traversal of the tree.
This means that the ordering of the nodes is determined based on where in the tree they appear - this is not the way a binary search tree typically works, where the nodes' elements have some ordering that's not dependent on where in the tree it appears (e.g. f is greater than a in a regular BST ordered lexicographically, but in our case f may be smaller or greater than a, since it's ordered based on the index of f and a).
To add or get, we start at the root and recursively go through the tree, determining whether our insert or lookup position is to the left or right based on the target index and the subtree sizes.
More specifically, we have the following recursive definitions:
(with some added complexity for null nodes and actually inserting the node)
node.add(index, element):
if index <= left.subtreeSize
left.add(index, element)
else
// anything to the right is after left subtree and current node, so those must be excluded
right.add(index - left.subtreeSize - 1, element)
node.get(index, element):
if index == left.subtreeSize
return node
if index < left.subtreeSize
return left.get(index)
else
return right.get(index - left.subtreeSize - 1)
To understand this better, the following example tree might be helpful:
Values: Indices (in-order pos): Subtree sizes:
a 5 8
/ \ / \ / \
b g 1 6 5 2
/ \ \ / \ \ / \ \
f c h 0 3 7 1 3 1
/ \ / \ / \
e d 2 4 1 1
If we want to insert a new node at position 5, for example, it will be inserted to the right of d.
Below is a small test program to demonstrate this (creating the tree shown above).
Note that balancing will still need to be done to achieve O(log n) running time per operation.
class Test
{
static class Node<T>
{
Node<T> left, right;
T data;
int subtreeCount;
Node(T data) { this.data = data; subtreeCount = 1; }
public String toString(int spaces, char leftRight)
{
return String.format("%" + spaces + "s%c: %s\n", "", leftRight, data.toString())
+ (left != null ? left.toString(spaces+3, 'L') : "")
+ (right != null ? right.toString(spaces+3, 'R') : "");
}
int subtreeSize(Node<T> node)
{
if (node == null)
return 0;
return node.subtreeCount;
}
// combined add and get into 1 function for simplicity
// if data is null, it is an get, otherwise it's an add
private T addGet(int index, T data)
{
if (data != null)
subtreeCount++;
if (index == subtreeSize(left) && data == null)
return this.data;
if (index <= subtreeSize(left))
{
if (left == null && data != null)
return (left = new Node<>(data)).data;
else
return left.addGet(index, data);
}
else if (right == null && data != null)
return (right = new Node<>(data)).data;
else
return right.addGet(index-subtreeSize(left)-1, data);
}
}
static class TreeArray<T>
{
private Node<T> root;
public int size() { return (root == null ? 0 : root.subtreeCount); }
void add(int index, T data)
{
if (index < 0 || index > size())
throw new IndexOutOfBoundsException("Index: " + index + ", Size: " + size());
if (root == null)
root = new Node<>(data);
else
root.addGet(index, data);
}
T get(int index)
{
if (index < 0 || index >= size())
throw new IndexOutOfBoundsException("Index: " + index + ", Size: " + size());
return root.addGet(index, null);
}
#Override
public String toString() { return root == null ? "Empty" : root.toString(1, 'X'); }
}
public static void main(String[] args)
{
TreeArray<String> tree = new TreeArray<>();
tree.add(0, "a");
tree.add(0, "b");
tree.add(1, "c");
tree.add(2, "d");
tree.add(1, "e");
tree.add(0, "f");
tree.add(6, "g");
tree.add(7, "h");
System.out.println("Tree view:");
System.out.print(tree);
System.out.println("Elements in order:");
for (int i = 0; i < tree.size(); i++)
System.out.println(i + ": " + tree.get(i));
}
}
This outputs:
Tree view:
X: a
L: b
L: f
R: c
L: e
R: d
R: g
R: h
Elements in order:
0: f
1: b
2: e
3: c
4: d
5: a
6: g
7: h
Live demo.
LinkedList is a linked-list with access\insert\remove requires O(n), linked-lists support sequential access O(n).
ArrayList is an array with insert\remove requires O(2n), but access requires O(1), arrays support random access O(1).
to find a more optimal hybrid structure, you can start with this:
template <T>
public class LinkedArrayList
{
LinkedList<ArrayList<T>> list;
public LinkedArrayList ()
{
list = new LinkedList<ArrayList<T>> ();
}
// ..
}
You'll have to balance segments (arrays) in the list between access complexity, and insert\remove complexity
Calculate the longest path between two nodes.
The path is in an arch.
Signature of method is:
public static int longestPath(Node n)
In the example binary tree below, it is 4 (going thru 2-3-13-5-2).
This is what I have right now and for the given tree it just returns 0.
public static int longestPath(Node n) {
if (n != null) {
longestPath(n, 0);
}
return 0;
}
private static int longestPath(Node n, int prevNodePath) {
if (n != null && n.getLeftSon() != null && n.getRightSon() != null) {
int currNodePath = countLeftNodes(n.getLeftSon()) + countRightNodes(n.getRightSon());
int leftLongestPath = countLeftNodes(n.getLeftSon().getLeftSon()) + countRightNodes(n.getLeftSon().getRightSon());
int rightLongestPath = countLeftNodes(n.getRightSon().getLeftSon()) + countRightNodes(n.getRightSon().getRightSon());
int longestPath = currNodePath > leftLongestPath ? currNodePath : leftLongestPath;
longestPath = longestPath > rightLongestPath ? longestPath : rightLongestPath;
longestPath(n.getLeftSon(), longestPath);
longestPath(n.getRightSon(), longestPath);
return longestPath > prevNodePath ? longestPath : prevNodePath;
}
return 0;
}
private static int countLeftNodes(Node n) {
if (n != null) {
return 1+ countLeftNodes(n.getLeftSon());
}
return 0;
}
private static int countRightNodes(Node n) {
if (n != null) {
return 1+ countRightNodes(n.getRightSon());
}
return 0;
}
I understand that I'm missing a key concept somewhere... My brain goes crazy when I try tracking the flow of execution...
Am I right by saying that by finding the longest path among the root, its left & right nodes and then recurse on its left & right nodes passing them the longest path from previous method invocation and finally (when?) return the longest path, I'm not certain as to how you go about returning it...
Maybe it is just as simple:
public static int longestPath(Node n) {
if (n != null) {
return longestPath(n, 0); // forgot return?
}
return 0;
}
Its more complicated than one might think at first sight. Consider the following tree:
1
/ \
2 3
/ \
4 5
/ \ \
6 7 8
/ \ \
9 a b
In this case, the root node is not even in the longest path (a-7-4-2-5-8-b).
So, what you must do is the following: For each node n you must compute the following:
compute longest path in left subtree starting with the root of the left subtree (called L)
compute longest path in right subtree starting with the root of the right subtree (called R)
compute the longest path in left subtree (not necessarily starting with the root of the left subtree) (called l)
compute the longest path in right subtree (not necessarily starting with the root of the right subtree) (called r)
Then, decide, which combination maximizes path length:
L+R+2, i.e. going from a subpath in left subtree to current node and from current node through a subpath in right subtree
l, i.e. just take the left subtree and exclude the current node (and thus right subtree) from path
r, i.e. just take the right subtree and exclude the current node (and thus left subtree) from path
So I would do a little hack and for every node not return just a single int, but a triple of integers containing (L+R+2, l, r). The caller then must decide what to do with this result according to the above rules.
A correct algorithm is:
Run DFS from any node to find the farthest leaf node. Label that node T.
Run another DFS to find the farthest node from T.
The path you found in step 2 is the longest path in the tree.
This algorithm will definitely work, and you're not limited to just binary trees either. I'm not sure about your algorithm:
Am I right by saying that by finding the longest path among the root, its left & right nodes and then recurse on its left & right nodes passing them the longest path from previous method invocation and finally (when???) return the longest path, I'm not certain as to how you go about returning it...
because I don't understand what exactly you're describing. Can you work it by hand on an example or try to explain it better? That way you might get better help understanding if it's correct or not.
You seem to be attempting a recursive implementation of basically the same thing just simplified for binary trees. Your code seems rather complicated for this problem however. Check the discussion here for a simpler implementation.
public int longestPath() {
int[] result = longestPath(root);
return result[0] > result[1] ? result[0] : result[1];
}
// int[] {self-contained, root-to-leaf}
private int[] longestPath(BinaryTreeNode n) {
if (n == null) {
return new int[] { 0, 0 };
}
int[] left = longestPath(n.left);
int[] right = longestPath(n.right);
return new int[] { Util.max(left[0], right[0], left[1] + right[1] + 1),
Util.max(left[1], right[1]) + 1 };
}
Simple Implementation:
int maxDepth(Node root) {
if(root == null) {
return 0;
} else {
int ldepth = maxDepth(root.left);
int rdepth = maxDepth(root.right);
return ldepth>rdepth ? ldepth+1 : rdepth+1;
}
}
int longestPath(Node root)
{
if (root == null)
return 0;
int ldepth = maxDepth(root.left);
int rdepth = maxDepth(root.right);
int lLongPath = longestPath(root.left);
int rLongPath = longestPath(root.right);
return max(ldepth + rdepth + 1, max(lLongPath, rLongPath));
}
Here is my recursive solution in C++:
int longest_dis(Node* root) {
int height1, height2;
if( root==NULL)
return 0;
if( root->left == NULL ) && ( root->right == NULL )
return 0;
height1 = height(root->left); // height(Node* node) returns the height of a tree rooted at node
height2 = height(root->right);
if( root->left != NULL ) && ( root->right == NULL )
return max(height1+1, longest_dis(root->left) );
if( root->left == NULL ) && ( root->right != NULL )
return max(height2+1, longest_dis(root->right) );
return max(height1+height2+2, longest_dis(root->left), longestdis(root->right) );
}
Taking into account #phimuemue example and #IVlad solution, I decided to check it out myself, so here is my implementation of #IVlad solution in python:
def longestPath(graph,start, path=[]):
nodes = {}
path=path+[start]
for node in graph[start]:
if node not in path:
deepestNode,maxdepth,maxpath = longestPath(graph,node,path)
nodes[node] = (deepestNode,maxdepth,maxpath)
maxdepth = -1
deepestNode = start
maxpath = []
for k,v in nodes.iteritems():
if v[1] > maxdepth:
deepestNode = v[0]
maxdepth = v[1]
maxpath = v[2]
return deepestNode,maxdepth +1,maxpath+[start]
if __name__ == '__main__':
graph = { '1' : ['2','3'],
'2' : ['1','4','5'],
'3' : ['1'],
'4' : ['2','6','7'],
'5' : ['2','8'],
'6' : ['4'],
'7' : ['4','9','a'],
'8' : ['5','b'],
'9' : ['7'],
'a' : ['7'],
'b' : ['8']
}
"""
1
/ \
2 3
/ \
4 5
/ \ \
6 7 8
/ \ \
9 a b
"""
deepestNode,maxdepth,maxpath = longestPath(graph,'1')
print longestPath(graph, deepestNode)
>>> ('9', 6, ['9', '7', '4', '2', '5', '8', 'b'])
I think You are overcomplicating things.
Think about the longest path that goes through the node n and doesn't go up to the parent of n. What is the relationship between the length of that path and the heights of both subtries connected to n?
After figuring that out, check the tree recursively reasoning like this:
The longest path for a subtree with the root n is the longest path of the following three:
The longest path in the subtree, whose root is n.left_child
The longest path in the subtree, whose root is n.right_child
The longest path, that goes through the node n and doesn't go up to the parent of n
What if, for each node n, your goal was to compute these two numbers:
f(n): The length of the longest path in the tree rooted at n
h(n): The height of the tree that is rooted at n.
For each terminal node (nodes having null left and right nodes), it is obvious that f and h are both 0.
Now, the h of each node n is:
0 if n.left and n.right are both null
1 + h(n.left) if only n.left is non-null
1 + h(n.right) if only n.right is non-null
1 + max(h(n.left), h(n.right)) if both n.left and n.right are non-null
And f(n) is:
0 if n.left and n.right are both null
max(f(n.left), h(n)) if only n.left is non-null
?? if only n.right is non-null
?? if both n.left and n.right are non-null
(You need to figure out what replaces the two "??" placeholders. There are choices that make this strategy work. I have tested it personally.)
Then, longestPath(Node n) is just f(n):
public class SO3124566
{
static class Node
{
Node left, right;
public Node()
{
this(null, null);
}
public Node(Node left, Node right)
{
this.left = left;
this.right = right;
}
}
static int h(Node n)
{
// ...
}
static int f(Node n)
{
// ...
}
public static int longestPath(Node n)
{
return f(n);
}
public static void main(String[] args)
{
{ // #phimuemue's example
Node n6 = new Node(),
n9 = new Node(),
a = new Node(),
n7 = new Node(n9, a),
n4 = new Node(n6, n7),
b = new Node(),
n8 = new Node(null, b),
n5 = new Node(null, n8),
n2 = new Node(n4, n5),
n3 = new Node(),
n1 = new Node(n2, n3);
assert(longestPath(n1) == 6);
}{ // #Daniel Trebbien's example: http://pastebin.org/360444
Node k = new Node(),
j = new Node(k, null),
g = new Node(),
h = new Node(),
f = new Node(g, h),
e = new Node(f, null),
d = new Node(e, null),
c = new Node(d, null),
i = new Node(),
b = new Node(c, i),
a = new Node(j, b);
assert(longestPath(a) == 8);
}
assert(false); // just to make sure that assertions are enabled.
// An `AssertionError` is expected on the previous line only.
}
}
You should be able to write recursive implementations of f and h to make this code work; however, this solution is horribly inefficient. Its purpose is just to understand the calculation.
To improve the efficiency, you could use memoization or convert this to a non-recursive calculation that uses stack(s).
Well, umm if I've understood your question correctly, here is my solution [but in C++(I'm sorry)]:
int h(const Node<T> *root)
{
if (!root)
return 0;
else
return max(1+h(root->left), 1+h(root->right));
}
void longestPath(const Node<T> *root, int &max)
{
if (!root)
return;
int current = h(root->left) + h(root->right) + 1;
if (current > max) {
max = current;
}
longestPath(root->left, max);
longestPath(root->right, max);
}
int longest()
{
int max = 0;
longestPath(root, max);
return max;
}