LeetCode（102） Binary Tree Level Order Traversal

题目

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:



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

OJ’s Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where ‘#’ signifies a path terminator where no node exists below.

Here’s an example:



The above binary tree is serialized as “{1,2,3,#,#,4,#,#,5}”.

分析

给定一颗二叉树，返回其层序遍历序列。

要求按照二叉树的结构，分层存储节点元素。

既然要求分层返回遍历结果，我们知道，层序遍历是利用队列先进先出的结构特点，依次遍历。按照本题要求，在遍历过程中必须将每层涉及到的节点单独存储，才能保证得到独立层节点序列。

//定义两个队列，一个存储所有的父节点,另一个存储他们的子节点也就是子层

AC代码

/**
 * 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) {
        //层次遍历，分层存储
        if (!root)
            return vector<vector<int> >();

        vector<vector<int> > ret;

        //定义两个队列，一个存储所有的父节点
        queue<TreeNode *> parents;

        parents.push(root);

        while (!parents.empty())
        {
            //存储当前层的遍历结果
            vector<int> tmp;
            //定义队列另一个存储他们的子节点也就是子层
            queue<TreeNode *> childs;
            while (!parents.empty())
            {
                TreeNode *node = parents.front();
                tmp.push_back(node->val);
                //弹出当前父节点
                parents.pop();

                if (node->left)
                    childs.push(node->left);

                if (node->right)
                    childs.push(node->right);
            }
            //存储当前层的遍历结果
            ret.push_back(tmp);

            //遍历下一层
            parents = childs;
        }
        return ret;
    }
};

GitHub测试程序源码