Cousins In Binary Tree Problem
Description
LeetCode Problem 993.
Given the root of a binary tree with unique values and the values of two different nodes of the tree x and y, return true if the nodes corresponding to the values x and y in the tree are cousins, or false otherwise.
Two nodes of a binary tree are cousins if they have the same depth with different parents. Note that in a binary tree, the root node is at the depth 0, and children of each depth k node are at the depth k + 1.
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
Constraints:
- The number of nodes in the tree is in the range [2, 100].
- 1 <= Node.val <= 100
- Each node has a unique value.
- x != y
- x and y are exist in the tree.
Sample C++ Code
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/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
void rec(TreeNode* root, int x, int y, int d, int parent) {
if (!root) return;
if (root->val == x)
dx = d, x_par = parent;
if (root->val == y)
dy = d, y_par = parent;
if (dx != -1 && dy != -1) return;
rec(root->left, x, y, d+1, root->val);
rec(root->right, x, y, d+1, root->val);
}
bool isCousins(TreeNode* root, int x, int y) {
rec(root, x, y, 0, -1);
return dx == dy && x_par != y_par;
}
private:
int dx = -1, dy = -1, x_par = -1, y_par = -1;
};