Binary Tree Inorder Traversal (leetcode)

题目：

Given a binary tree, return the inorder traversal of its nodes' values.

For example:

Given binary tree

{1,#,2,3}

,

1
\
2
/
3

return

[1,3,2]

.

Note: Recursive solution is trivial, could you do it iteratively?

confused what

"{1,#,2,3}"

means? >
read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:

1
/ \
2   3
/
4
\
5

The above binary tree is serialized as

"{1,2,3,#,#,4,#,#,5}"

. https://oj.leetcode.com/problems/binary-tree-inorder-traversal/ href="https://oj.leetcode.com/problems/binary-tree-inorder-traversal/" target=_blank>点击打开链接

二叉树的中序遍历，其中inorder函数是递归版本的遍历，而inorderIterat是迭代版的中序遍历。

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

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

void inorder(TreeNode *root,vector<int> &result)
{
if(root==NULL)
return ;
inorder(root->left,result);
result.push_back(root->val);
inorder(root->right,result);
}

void inorderIterat(TreeNode *root,vector<int> &result)
{
stack<TreeNode*> s;
s.push(root);
TreeNode* temp=s.top();
while(temp->left!=NULL)
{
temp=temp->left;
s.push(temp);
}
while(!s.empty())
{
temp=s.top();
result.push_back(temp->val);
s.pop();
if(temp->right!=NULL)
{
s.push(temp->right);
temp=s.top();
while(temp->left!=NULL)
{
temp=temp->left;
s.push(temp);
}
}
}
}

vector<int> inorderTraversal(TreeNode *root)
{
vector<int> result;
if(root==NULL)
return result;
//	inorder(root,result);//递归版本
inorderIterat(root,result);//迭代版本
return result;
}

int main()
{
TreeNode *root=new TreeNode(1);
root->left=new TreeNode(2);
root->right=new TreeNode(3);
vector<int> result=inorderTraversal(root);
for(int i=0;i<result.size();i++)
cout<<result[i]<<" ";
cout<<endl;

system("pause");
return 0;
}