leetcode_107题——Binary Tree Level Order Traversal II （二叉树，广度优先搜索，队列，栈）

Binary Tree Level Order Traversal II

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Question Solution

Given a binary tree, return the bottom-up level order traversal of its nodes‘ values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3
   /   9  20
    /     15   7

return its bottom-up level order traversal as:

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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

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

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

/*采用广度优先搜索的方法，一层一层的去搜索，搜索时用到了队列，将每一层的结点进队列，然后将每一层
的一次输出，而在题目要求中反过来显示，所以又增加了一个堆栈，来为后面反过来显示节点数据*/
vector<vector<int> > levelOrder(TreeNode *root) {
	vector<vector<int> > last_result;//最终的结果
	stack<vector<int> >  stack_result;//作为中间结果
	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();
		stack_result.push(temp_result);//将每一层的数据压栈
	}
	while(!stack_result.empty())//最后按栈中的元素再反向的输出
	{
		last_result.push_back(stack_result.top());
		stack_result.pop();
	}
	return last_result;
}
int main()
{

}