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()
{

}