[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 atroot
.
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 }