leetcode 094 Binary Tree Inorder Traversal

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}"

.

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class Solution {
public:
void inOrder(TreeNode* root, vector<int> &res) {
if(root!=NULL) {
inOrder(root->left, res);
res.push_back(root->val);
inOrder(root->right, res);
}
return;
}
vector<int> inorderTraversal(TreeNode* root) {
vector<int> res;
stack<TreeNode*> s;
if(root==NULL) return res;

s.push(root);

while(!s.empty()) {
TreeNode *temp = s.top();

if(temp->left!=NULL) {
s.push(temp->left);
temp->left=NULL;
} else {
s.pop();
res.push_back(temp->val);
if(temp->right!=NULL) {
s.push(temp->right);
}
}
}
return res;
}
}