[LeetCode] 1602. Find Nearest Right Node in Binary Tree

Given the root of a binary tree and a node u in the tree, return the nearest node on the same level that is to the right of u, or return null if u is the rightmost node in its level.

Example 1:

Input: root = [1,2,3,null,4,5,6], u = 4
Output: 5
Explanation: The nearest node on the same level to the right of node 4 is node 5.

Example 2:

Input: root = [3,null,4,2], u = 2
Output: null
Explanation: There are no nodes to the right of 2.

Example 3:

Input: root = [1], u = 1
Output: null

Example 4:

Input: root = [3,4,2,null,null,null,1], u = 4
Output: 2

Constraints:

The number of nodes in the tree is in the range [1, 105].
1 <= Node.val <= 105
All values in the tree are distinct.
u is a node in the binary tree rooted at root.

找到二叉树中最近的右侧节点。

题目很简单，就是常规的BFS，唯一需要注意的是如果在当前层存在u节点的话，需要判断u节点是否是当前层从左往右的最后一个节点，如果不是，他才会有右侧节点。

时间O(n)

空间O(n)

Java实现

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode() {}
 8  *     TreeNode(int val) { this.val = val; }
 9  *     TreeNode(int val, TreeNode left, TreeNode right) {
10  *         this.val = val;
11  *         this.left = left;
12  *         this.right = right;
13  *     }
14  * }
15  */
16 class Solution {
17     public TreeNode findNearestRightNode(TreeNode root, TreeNode u) {
18         Queue<TreeNode> queue = new LinkedList<>();
19         queue.offer(root);
20         while (!queue.isEmpty()) {
21             int size = queue.size();
22             for (int i = 0; i < size; i++) {
23                 TreeNode cur = queue.poll();
24                 if (cur.val == u.val) {
25                     if (i < size - 1) {
26                         return queue.poll();
27                     } else {
28                         return null;
29                     }
30                 }
31                 if (cur.left != null) {
32                     queue.offer(cur.left);
33                 }
34                 if (cur.right != null) {
35                     queue.offer(cur.right);
36                 }
37             }
38         }
39         return null;
40     }
41 }

LeetCode 题目总结