LeetCode 993. Cousins in Binary Tree

原题链接在这里：https://leetcode.com/problems/cousins-in-binary-tree/

题目：

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.

题解：

Perform level order tranersal.

When polling the current node, if its left and right child are not null and they equal to x and y, then x and y are sibling, not cousins, return false.

Have two boolean value mark if x is found and y is found. If bouth found on the same level, return true.

Time Complexity: O(n).

Space: O(n).

AC Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isCousins(TreeNode root, int x, int y) {
        if(root == null){
            return false;
        }
        
        LinkedList<TreeNode> que = new LinkedList<>();
        que.add(root);
        
        while(!que.isEmpty()){
            int size = que.size();
            boolean xFound = false;
            boolean yFound = false;
            
            while(size-- > 0){
                TreeNode cur = que.poll();
                if(cur.val == x){
                    xFound = true;
                }
                
                if(cur.val == y){
                    yFound = true;
                }
                
                if(cur.left != null && cur.right != null){
                    if(cur.left.val == x && cur.right.val == y){
                        return false;
                    }
                    
                    if(cur.left.val == y && cur.right.val == x){
                        return false;
                    }
                }
                
                if(cur.left != null){
                    que.add(cur.left);
                }
                
                if(cur.right != null){
                    que.add(cur.right);
                }
            }
            
            if(xFound && yFound){
                return true;
            }
        }
        
        return false;
    }
}