README.md

1602. 找到二叉树中最近的右侧节点

English Version

题目描述

给定一棵二叉树的根节点 root 和树中的一个节点 u ，返回与 u 所在层中距离最近的右侧节点，当 u 是所在层中最右侧的节点，返回 null 。

 

示例 1：

输入：root = [1,2,3,null,4,5,6], u = 4
输出：5
解释：节点 4 所在层中，最近的右侧节点是节点 5。

示例 2：

输入：root = [3,null,4,2], u = 2
输出：null
解释：2 的右侧没有节点。

示例 3：

输入：root = [1], u = 1
输出：null

示例 4：

输入：root = [3,4,2,null,null,null,1], u = 4
输出：2

 

提示:

• 树中节点个数的范围是 [1, 105] 。
• 1 <= Node.val <= 105
• 树中所有节点的值是唯一的。
• u 是以 root 为根的二叉树的一个节点。

解法

BFS 层序遍历。

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def findNearestRightNode(self, root: TreeNode, u: TreeNode) -> TreeNode:
        q = deque([root])
        while q:
            n = len(q)
            for i in range(n):
                node = q.popleft()
                if node == u:
                    return None if i == n - 1 else q.popleft()
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode findNearestRightNode(TreeNode root, TreeNode u) {
        Deque<TreeNode> q = new ArrayDeque<>();
        q.offer(root);
        while (!q.isEmpty()) {
            for (int i = 0, n = q.size(); i < n; ++i) {
                TreeNode node = q.poll();
                if (node == u) {
                    return i == n - 1 ? null : q.poll();
                }
                if (node.left != null) {
                    q.offer(node.left);
                }
                if (node.right != null) {
                    q.offer(node.right);
                }
            }
        }
        return null;
    }
}

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* findNearestRightNode(TreeNode* root, TreeNode* u) {
        queue<TreeNode*> q;
        q.push(root);
        while (!q.empty())
        {
            for (int i = 0, n = q.size(); i < n; ++i)
            {
                TreeNode* node = q.front();
                q.pop();
                if (node == u) return i == n - 1 ? nullptr : q.front();
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }
        }
        return nullptr;
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func findNearestRightNode(root *TreeNode, u *TreeNode) *TreeNode {
	q := []*TreeNode{root}
	for len(q) > 0 {
		t := q
		q = nil
		for i, node := range t {
			if node == u {
				if i == len(t)-1 {
					return nil
				}
				return t[i+1]
			}
			if node.Left != nil {
				q = append(q, node.Left)
			}
			if node.Right != nil {
				q = append(q, node.Right)
			}
		}
	}
	return nil
}

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