算法作业HW18：LeetCode102 Binary Tree Level Order Traversal

Description:

  Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

Note:

For example:
Given binary tree [3,9,20,null,null,15,7],

    3   / \  9  20    /  \   15   7

return its level order traversal as:

[  [3],  [9,20],  [15,7]]

 

Solution:

  Analysis and Thinking:

这是一道明显的层序遍历二叉树的题目，遵循一般的思路即：如果队列不空，将树的根入队列饿，再进行循环

  Steps:

1.初始化队列，输入树

2.将树的根节点入队，并进行循环

3.当队列不空，将队首部的元素出队，并判断其是否有左、右子树，分别将其子树入队

4.遍历完成，返回结果

 

Codes:

/**  * 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> > result;                 if(!root) 

    return res;  

        queue<TreeNode *> tempQue;          
tempQue.push(root);  
 while(!tempQue.empty()){ vector<int> tempLevel; int size = tempQue.size(); for(int i = 0; i < size; i ++){ TreeNode *t = tempQue.front(); tempQue.pop(); tempLevel.push_back(t -> val); if(t -> left) tempQue.push(t -> left); if(t -> right) tempQue.push(t -> right); } result.push_back(everylevel); } return result; } };



 
Results: