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.
/**
* 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)
{
vector<int> res;
get_res(root,res);
return res;
}
void get_res(TreeNode* x,vector<int>& res)
{
if(x == NULL)
return;
get_res(x->left,res);
res.push_back(x->val);
get_res(x->right,res);
}
};LeetCode--Binary Tree Inorder Traversal
原文:http://blog.csdn.net/shaya118/article/details/42680603