[Leetcode] Binary Tree Level Order Traversal

最新推荐文章于 2025-09-18 03:57:00 发布

原创最新推荐文章于 2025-09-18 03:57:00 发布 · 396 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#BFS #Queue

Leetcode 专栏收录该内容

67 篇文章

订阅专栏

本文介绍了一种经典的二叉树层次遍历算法，通过使用队列数据结构并插入NULL标志位来实现从左到右、按层遍历二叉树节点的值。通过实例展示了如何对给定的二叉树进行层次遍历，并返回每层节点值。

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.

经典方法，用Queue的data structure，插入NULL的标志位。

/**
 * Definition for binary tree
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        vector<vector<int> > result;
        if(root==NULL) return result;
        queue<TreeNode*> q;
        
        q.push(root);
        q.push(NULL);
        
        vector<int> levelVal;
        while(true)
        {
            TreeNode* cur = q.front();
            q.pop();
            
            if(cur==NULL)
            {
                result.push_back(levelVal);
                levelVal.clear();
                if(q.size()==0)
                {
                    break;
                }
                else
                {
                    q.push(NULL);
                }
            }
            else
            {
                levelVal.push_back(cur->val);
                if(cur->left) q.push(cur->left);
                if(cur->right) q.push(cur->right);
            }
        }
        return result;
    }
};