993. Cousins in Binary Tree

#### QUESTION:

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`.

#### EXPLANATION:

1.需要判断两个值在不在一个level

2.需要判断是不是一个root

#### SOLUTION:

``````/**
* 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) {
Queue<TreeNode> q1 = new ArrayDeque();
Queue<TreeNode> q2 = new ArrayDeque();
int turn = 0;
int[][] indexX = new int[1][2];
int[][] indexY = new int[1][2];
while (!q1.isEmpty() || !q2.isEmpty()){
Queue<TreeNode> outQueue = turn%2==0 ? q1: q2;
Queue<TreeNode> inQueue = turn%2==0 ? q2:q1;
while (!outQueue.isEmpty()){
TreeNode poll = outQueue.poll();
if((poll.left!=null && poll.left.val == x )|| (poll.right!=null && poll.right.val==x)){
indexX[0][0] = turn+1;
indexX[0][1] = poll.val;
}
if((poll.left!=null && poll.left.val == y )|| (poll.right!=null && poll.right.val==y)){
indexY[0][0] = turn+1;
indexY[0][1] = poll.val;
}