LEETCODE: Binary Tree Level Order Traversal

停止更新+1

于 2014-12-30 13:02:58 发布

阅读量287

点赞数

CC 4.0 BY-SA版权

文章标签： LEETCODE 算法二叉树遍历

本文链接：https://blog.youkuaiyun.com/anyicheng2015/article/details/42264721

LEETCODE 专栏收录该内容

153 篇文章

订阅专栏

本文介绍了一种解决二叉树层次遍历问题的方法，通过使用两个队列来实现节点值从左到右、逐层返回的功能。示例中详细解释了如何处理每个层级的节点及其子节点。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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]
]

我用了两个队列的方式来做这题。

/**
 * 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) {
        vector<vector<int> > results;
        if(root == NULL) return results;
        queue<TreeNode *> onelevel;
        onelevel.push(root);
        while(!onelevel.empty()) {
            // Get a level and get TreeNode of the next level.
            vector<int> valOfCurrentLevel;
            queue<TreeNode *> nextlevel;
            while(!onelevel.empty()) {
                TreeNode *front = onelevel.front();
                valOfCurrentLevel.push_back(front->val);
                if(front->left != NULL)
                    nextlevel.push(front->left);
                if(front->right != NULL)
                    nextlevel.push(front->right);
                onelevel.pop();
            }
            results.push_back(valOfCurrentLevel);
            onelevel = nextlevel;
        }
        return results;
    }
};