leetcode_102题——Binary Tree Level Order Traversal（二叉树，广度优先搜索，队列）

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.

Hide Tags

Tree Breadth-first Search

Have you met this question in a real interview?

Yes

No

Discuss

#include<iostream>
#include<vector>
#include<list>
using namespace std;

//Definition for binary tree
struct TreeNode {
      int val;
      TreeNode *left;
      TreeNode *right;
      TreeNode(int x) : val(x), left(NULL), right(NULL) {}
  };

vector<vector<int> > levelOrder(TreeNode *root) {
	    vector<vector<int> > last_result;//最终的结果
        if(root==NULL)
			return last_result;

		list<TreeNode*> temp;//做计算的队列
		TreeNode* temp_node;//作为中间变量
		//int depth=1;
		int row_size=1;//记录每一层的结点个数
		temp.push_back(root);

		while(!temp.empty())
		{
			vector<int> temp_result;//设置一个装每一层结点的中间vector
			while(row_size--)
			{
				temp_node=temp.front();
				temp.pop_front();
				temp_result.push_back(temp_node->val);
				if(temp_node->left!=NULL)//将左子树进队列
					temp.push_back(temp_node->left);
				if(temp_node->right!=NULL)//将右子树进队列
					temp.push_back(temp_node->right);
			}
			row_size=temp.size();
			last_result.push_back(temp_result);
		}
		return last_result;
    }

int main()
{

}