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## 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?

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}`".

## Solution

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

```void loop(vector<int> &result, TreeNode *root) {
if (root == nullptr)
return;
stack<TreeNode *> s;
TreeNode *cur = root;
while (cur || !s.empty()) {
while (cur) {
s.push(cur);
cur = cur->left;
}
cur = s.top();
s.pop();
result.push_back(cur->val);
cur = cur->right;
}
}```
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