描述：
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}"
代码：1A
/**
* Definition for binary tree
* 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) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector<int> res;
if(NULL == root) return res;
traversal(root->left,res);
res.push_back(root->val);
traversal(root->right,res);
return res;
}
void traversal(TreeNode *root,vector<int> &res)
{
if(!root) return;
traversal(root->left,res);
res.push_back(root->val);
traversal(root->right,res);
}
};
递归就是简洁，如果换成迭代.....