算法作業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:

這是一道明顯的層序遍歷二叉樹的題目，遵循一般的思路即：如果隊列不空，將樹的根入隊列餓，再進行循環

  Steps:

1.初始化隊列，輸入樹

2.將樹的根節點入隊，並進行循環

3.當隊列不空，將隊首部的元素出隊，並判斷其是否有左、右子樹，分別將其子樹入隊

4.遍歷完成，返回結果

 

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: