[LeetCode]Binary Tree Level Order Traversal

二叉树层次遍历

最新推荐文章于 2025-09-07 20:32:33 发布

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最新推荐文章于 2025-09-07 20:32:33 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/skycore/p/5004692.html

本文介绍了如何使用广度优先遍历的方法解决二叉树的层次遍历问题，并提供了递归版和迭代版两种实现方式的代码示例。

题目描述:(链接)

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree {3,9,20,#,#,15,7},

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

解题思路：

广度优先遍历

递归版：

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     vector<vector<int>> levelOrder(TreeNode* root) {
13         travel(root, 1);
14         return result;
15     }
16     
17     void travel(TreeNode *root, int level) {
18         if (!root) return;
19         if (level > result.size()) {
20             result.push_back(vector<int>());
21         }
22         
23         result[level - 1].push_back(root->val);
24         travel(root->left, level + 1);
25         travel(root->right, level + 1);
26     }
27 private:
28     vector<vector<int>> result;
29 };

迭代版:

借助一个队列实现先进先出：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> result;
        if (!root) return result;
        
        queue<TreeNode *> current, next;
        vector<int> level;
        current.push(root);
        
        while (!current.empty()) {
            while (!current.empty()) {
                TreeNode *tmp = current.front();
                current.pop();
                level.push_back(tmp->val);
                
                if (tmp->left != nullptr) next.push(tmp->left);
                if (tmp->right != nullptr) next.push(tmp->right);
            }
            
            result.push_back(level);
            level.clear();
            swap(current, next);
        }
        return result;
    }
};