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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.
Tree Breadth-first Search
#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()
{
}
leetcode_102题——Binary Tree Level Order Traversal(二叉树,广度优先搜索,队列)
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原文地址:http://www.cnblogs.com/yanliang12138/p/4431391.html