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;
    }
};