leetcode_103题——Binary Tree Zigzag Level Order Traversal（广度优先搜索，队列queue，栈stack）

Binary Tree Zigzag Level Order Traversal

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Question Solution

Given a binary tree, return the zigzag level order traversal of its nodes‘ values. (ie, from left to right, then right to left for the next level and alternate between).

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

    3
   /   9  20
    /     15   7

return its zigzag level order traversal as:

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

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

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#include<iostream>
#include<queue>
#include<stack>
using namespace std;

 struct TreeNode {
	     int val;
	     TreeNode *left;
	     TreeNode *right;
	     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
	 };

 /*这道题采用的是广度优先搜索的算法，对着二叉树一层一层的遍历，而对于题目中的要求，需要在先向右
 又接着向左遍历，所以应用了栈，在偶数行时，先将结点压入栈中，再依次输入到vector中
 */
vector<vector<int> > zigzagLevelOrder(TreeNode *root)
{
	vector<vector<int> > last_result;//最终的存放结果的vector
	if(root==NULL)
		return last_result;

	TreeNode* temp_node;
	queue<TreeNode*> temp;
	int row_size=1;//每一层结点的个数
	temp.push(root);
	int deepth=1;//层数

	while(!temp.empty())
	{
		vector<int> temp_relust;
		stack<int> temp_stack;
		while(row_size--)//每一层遍历
		{
			temp_node=temp.front();
			temp.pop();
			if(temp_node->left!=NULL)
				temp.push(temp_node->left);
			if(temp_node->right!=NULL)
				temp.push((temp_node->right));
			if(deepth%2==0)//偶数层的压栈
			{
				temp_stack.push(temp_node->val);
			}
			else
				temp_relust.push_back(temp_node->val);
		}
		if(deepth%2==0)//偶数层的出栈
		{
			while(!temp_stack.empty())
			{
			  temp_relust.push_back(temp_stack.top());
			  temp_stack.pop();
			}
		}
		row_size=temp.size();
		last_result.push_back(temp_relust);
		deepth++;
	}
	return last_result;
 }
int main()
{

}