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:
- The number of nodes in the tree will be between
2
and100
. - Each node has a unique integer value from
1
to100
.
level order traversal,用两个flag分别表示X、Y是否存在:hasX, hasY
每一层开始时把hasX, hasY重置,如果这一层同时存在X、Y,还要判断它们是否是同一个parent,如果不是则继续traversal。如果遍历一层后,X、Y同时存在,且不是同一个parent,返回true
time = O(n), space = O(n) worst case
/** * 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; } Queue<TreeNode> q = new LinkedList<>(); q.offer(root); while(!q.isEmpty()) { int size = q.size(); boolean hasX = false, hasY = false; for(int i = 0; i < size; i++) { TreeNode cur = q.poll(); if(cur.val == x) { hasX = true; } if(cur.val == y) { hasY = 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) { q.offer(cur.left); } if(cur.right != null) { q.offer(cur.right); } } if(hasX && hasY) { return true; } } return false; } }