# Description

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].

The original problem is here.

The original code is here.

# My Solution

I solve this problem in C++, as below:

``````/*
*Binary Tree Inorder Traversal
*Author: shuaijiang
*Email: zhaoshuaijiang8@gmail.com
*/
#include<iostream>
#include<stack>
#include<stdlib.h>
using namespace std;

/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> inorderTraversal(TreeNode* root) {
stack<TreeNode*> myStack;
vector<int> result;
if(root == NULL)
return result;
myStack.push(root);
while(!myStack.empty()){
TreeNode * node = myStack.top();
myStack.pop();
if(node->left == NULL && node->right == NULL)
result.push_back(node->val);
else{
TreeNode * left = NULL;
if(node->right != NULL){
myStack.push(node->right);
node->right = NULL;
}
if(node->left != NULL){
left = node->left;
node->left = NULL;
}
myStack.push(node);
if(left != NULL)
myStack.push(left);
}
}
return result;
}
};
``````

# Note

To solve the problem, I use a stack to save the node of the tree.