62-Binary Tree Level Order Traversal

1. Binary Tree Level Order Traversal My Submissions QuestionEditorial Solution
   Total Accepted: 102531 Total Submissions: 312211 Difficulty: Easy
   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]
]

思路：用队列存储每一层的节点，出队时，将下一层入新队，依次遍历每层
时间O(n),空间O(n)

/** * 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>> res;        if(root==NULL)return res;  //root为空的情况先处理，也可在while条件中加限制        queue<TreeNode*> levelnode;        vector<int> row;        levelnode.push(root);        while(!levelnode.empty()){            queue<TreeNode*> pre;            vector<int> tmp;            while(!levelnode.empty()){                TreeNode * treetmp=levelnode.front();                levelnode.pop();                tmp.push_back(treetmp->val);                if(treetmp->left!=NULL)pre.push(treetmp->left);                if(treetmp->right!=NULL)pre.push(treetmp->right);            }            res.push_back(tmp);            levelnode = pre;        }        return res;    }};