75 lines
2.0 KiB
C++
75 lines
2.0 KiB
C++
// Source : https://leetcode.com/problems/cousins-in-binary-tree/
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// Author : Hao Chen
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// Date : 2019-04-30
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/*****************************************************************************************************
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*
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* In a binary tree, the root node is at depth 0, and children of each depth k node are at depth k+1.
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*
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* Two nodes of a binary tree are cousins if they have the same depth, but have different parents.
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*
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* We are given the root of a binary tree with unique values, and the values x and y of two different
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* nodes in the tree.
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*
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* Return true if and only if the nodes corresponding to the values x and y are cousins.
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*
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* Example 1:
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*
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* Input: root = [1,2,3,4], x = 4, y = 3
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* Output: false
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*
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* Example 2:
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*
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* Input: root = [1,2,3,null,4,null,5], x = 5, y = 4
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* Output: true
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*
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* Example 3:
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*
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* Input: root = [1,2,3,null,4], x = 2, y = 3
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* Output: false
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*
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* Note:
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*
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* The number of nodes in the tree will be between 2 and 100.
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* Each node has a unique integer value from 1 to 100.
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*
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******************************************************************************************************/
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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* };
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*/
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class Solution {
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public:
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bool isCousins(TreeNode* root, int x, int y) {
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int dx=0, dy=0;
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TreeNode *px=root, *py=root;
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dx = DepthAndParent(root, px, 0, x);
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dy = DepthAndParent(root, py, 0, y);
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if (dx && dy){
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return (dx == dy && px != py);
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}
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return false;
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}
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int DepthAndParent(TreeNode* root, TreeNode*& parent, int depth, int x) {
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if (!root) return 0;
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if ( root->val == x) return depth;
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int d=0;
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parent = root;
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if ( ( d = DepthAndParent(root->left, parent, depth+1, x)) > 0 ) return d;
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parent = root;
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if ( ( d = DepthAndParent(root->right, parent, depth+1, x)) > 0 ) return d;
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return 0;
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}
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};
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