Cousins in Binary Tree
Problem Statement - link #
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.
Examples #
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
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 != yxandyare exist in the tree.
Solutions #
/**
* 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:
int xd,yd ;
TreeNode *xp,*yp;
void rc(TreeNode* root, TreeNode* p, int x, int y, int d){
if(!root) return;
if(x == root->val){
xd = d;
xp = p;
}
else if(y == root->val){
yd = d;
yp = p;
}
rc(root->left, root, x, y, d+1);
rc(root->right, root, x, y, d+1);
}
bool isCousins(TreeNode* root, int x, int y) {
rc(root, NULL, x, y,0);
return xd==yd and xp!=yp;
}
};