LeetCode每日一题——993. Cousins in Binary Tree

一、题目

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:

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 != y
x and y are exist in the tree.

二、题解

/**
 * 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:
    bool isCousins(TreeNode* root, int x, int y) {
        TreeNode* xf = nullptr;
        TreeNode* yf = nullptr;
        queue<TreeNode*> q;
        q.push(root);
        while(!q.empty()){
            int size = q.size();
            vector<TreeNode*> res;
            while(size--){
                TreeNode* t = q.front();
                q.pop();
                if(t->left){
                    q.push(t->left);
                    xf = t->left->val == x ? t:xf;
                    yf = t->left->val == y ? t:yf;
                }
                if(t->right){
                    q.push(t->right);
                    xf = t->right->val == x ? t:xf;
                    yf = t->right->val == y ? t:yf;
                }
            }
            if(xf && yf && (xf != yf)) return true;
            if(xf || yf) return false;
        }
        return false;
    }
};