# Binary Tree Inorder Traversal【94】

Datetime:2016-08-23 01:22:02         Topic: LeetCode          Share        Original >>

### 94. Binary Tree Inorder Traversal

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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.

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```/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
//思路首先：递归解法，烂大街的解法
class Solution {
public:
vector<int> inorderTraversal(TreeNode* root) {
if(root){
inorderTraversal(root->left);
result.push_back(root->val);
inorderTraversal(root->right);
}
return result;
}
private:
vector<int> result;
};```

```/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
//思路首先：中序式深度优先搜索了.....
class Solution {
public:
void dfs(TreeNode *root, vector<int> &result) {
if(NULL == root)
return;
dfs(root->left, result);
result.push_back(root->val);
dfs(root->right, result);
}
vector<int> inorderTraversal(TreeNode* root) {
dfs(root, result);
return result;
}
private:
vector<int> result;
};```