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题目LeetCode102

给你二叉树的根节点root，返回其节点值的层序遍历。 （即逐层地，从左到右访问所有节点）。

输入：root = [3,9,20,null,null,15,7]
输出：[[3],[9,20],[15,7]]

Python解法

代码示例（广度优先搜索）

from collections import deque from typing import List, Optional # 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 levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]: if not root: return [] res = [] q = deque([root]) # 队列初始：[3] while q: # 只要队列不为空 level = [] size = len(q) # 记录当前层有多少节点 for _ in range(size): node = q.popleft() # 取出队首 level.append(node.val) if node.left: q.append(node.left) if node.right: q.append(node.right) res.append(level) return res

过程展示

Java解法

代码示例（广度优先搜索）

import java.util.List; import java.util.ArrayList; import java.util.Queue; import java.util.LinkedList; 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 List<List<Integer>> levelOrder(TreeNode root) { List<List<Integer>> res = new ArrayList<>(); if (root == null) return res; Queue<TreeNode> queue = new LinkedList<>(); queue.offer(root); while (!queue.isEmpty()) { List<Integer> level = new ArrayList<>(); int size = queue.size(); for (int i = 0; i < size; i++) { TreeNode node = queue.poll(); level.add(node.val); if (node.left != null) queue.offer(node.left); if (node.right != null) queue.offer(node.right); } res.add(level); } return res; } }

C++解法

代码示例（广度优先搜索）

#include <vector> #include <queue> using namespace std; // 二叉树节点定义 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: vector<vector<int>> levelOrder(TreeNode* root) { vector<vector<int>> res; if (!root) return res; queue<TreeNode*> q; q.push(root); while (!q.empty()) { vector<int> level; int size = q.size(); for (int i = 0; i < size; ++i) { TreeNode* node = q.front(); q.pop(); level.push_back(node->val); if (node->left) q.push(node->left); if (node->right) q.push(node->right); } res.push_back(level); } return res; } };