Leetcode--Binary Tree Level Order Traversal

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},

    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.

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:

   1  / \ 2   3    /   4    \     5

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

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 Tree Breadth-first Search

/** * 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) {//二叉树的广度优先遍历又称按层次遍历。算法借助队列实现        vector<vector<int>>  res;        if(root==NULL)            return res;               // stack<TreeNode *> cur;        deque<TreeNode*> que;        que.push_back(root);        while(!que.empty())        {            vector<int> temp;            deque<TreeNode*> cur;            while(!que.empty())            {                temp.push_back(que.front()->val);                if(que.front()->left!=NULL)                    cur.push_back(que.front()->left);                if(que.front()->right!=NULL)                    cur.push_back(que.front()->right);                que.pop_front();            }            res.push_back(temp);                        que.assign(cur.begin(),cur.end());        }                return res;            }};