Cousins in Binary Tree
In a binary tree, the root node is at depth 0, and children of each depth k node are at depth k+1.
Two nodes of a binary tree are cousins if they have the same depth, but have different parents.
We are given the root of a binary tree with unique values, and the values x and y of two different nodes in the tree.
Return true if and only if the nodes corresponding to the values x and y are cousins.
Example 1:
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Input: root = [1,2,3,4], x = 4, y = 3
Output: false
Example 2:
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Input: root = [1,2,3,null,4,null,5], x = 5, y = 4
Output: true
Example 3:
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Input: root = [1,2,3,null,4], x = 2, y = 3
Output: false
Note:
The number of nodes in the tree will be between 2 and 100. Each node has a unique integer value from 1 to 100.
Java Solution:
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/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public boolean isCousins(TreeNode root, int x, int y) {
TreeNode ParentFirst = getParentWithChild(root, x);
TreeNode ParentSecond = getParentWithChild(root, y);
if(ParentFirst == null || ParentSecond == null){
return false;
}
if((ParentFirst.val != ParentSecond.val) &&
getDepth(root,ParentFirst) == getDepth(root,ParentSecond)){
return true;
}
else{
return false;
}
}
int getDepth(TreeNode root,TreeNode target){
if(root == null){
return -1;
}
if(root.val == target.val){
return 0;
}
int leftDepth = getDepth(root.left,target);
int righttDepth = getDepth(root.right,target);
if(leftDepth != -1){
return leftDepth + 1;
}
if(righttDepth != -1){
return righttDepth + 1;
}
return -1;
}
TreeNode getParentWithChild(TreeNode root, int x){
if(root == null){ return null;}
if((root.left != null && root.left.val == x) ||
(root.right!= null && root.right.val == x)){
return root;
}
TreeNode left = getParentWithChild(root.left,x);
TreeNode right = getParentWithChild(root.right,x);
if(left != null){
return left;
}
if(right != null){
return right;
}
return null;
}
}
