993. Cousins in Binary Tree

993. Cousins in Binary Tree (Easy)

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:

1. The number of nodes in the tree will be between 2 and 100.
2. Each node has a unique integer value from 1 to 100.

 

Related Topics:
Tree, Breadth-first Search

Similar Questions:

• Binary Tree Level Order Traversal (Medium)

Solution 1.

// OJ: https://leetcode.com/problems/cousins-in-binary-tree/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    bool isCousins(TreeNode* root, int x, int y) {
        if (root->val == x || root->val == y) return false;
        queue<pair<TreeNode*, TreeNode*>> q;
        q.emplace((TreeNode*)NULL, root);
        while (q.size()) {
            int cnt = q.size();
            TreeNode *a = NULL, *b = NULL;
            while (cnt--) {
                auto p = q.front();
                q.pop();
                if (p.second->val == x) a = p.first;
                if (p.second->val == y) b = p.first;
                if (p.second->left) q.emplace(p.second, p.second->left);
                if (p.second->right) q.emplace(p.second, p.second->right);
            }
            if (a || b) return a && b && a != b;
        }
        return false;
    }
};