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CousinsInBinaryTree.cpp
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CousinsInBinaryTree.cpp
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// Source : https://leetcode.com/problems/cousins-in-binary-tree/
// Author : Hao Chen
// Date : 2019-04-30
/*****************************************************************************************************
*
* 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:
*
* Input: root = [1,2,3,4], x = 4, y = 3
* Output: false
*
* Example 2:
*
* Input: root = [1,2,3,null,4,null,5], x = 5, y = 4
* Output: true
*
* Example 3:
*
* 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.
*
******************************************************************************************************/
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool isCousins(TreeNode* root, int x, int y) {
int dx=0, dy=0;
TreeNode *px=root, *py=root;
dx = DepthAndParent(root, px, 0, x);
dy = DepthAndParent(root, py, 0, y);
if (dx && dy){
return (dx == dy && px != py);
}
return false;
}
int DepthAndParent(TreeNode* root, TreeNode*& parent, int depth, int x) {
if (!root) return 0;
if ( root->val == x) return depth;
int d=0;
parent = root;
if ( ( d = DepthAndParent(root->left, parent, depth+1, x)) > 0 ) return d;
parent = root;
if ( ( d = DepthAndParent(root->right, parent, depth+1, x)) > 0 ) return d;
return 0;
}
};