94. Binary Tree Inorder Traversal

Given a binary tree, return the inorder traversal of its nodes' values.

Example:

Input: [1,null,2,3]
   1
    \
     2
    /
   3

Output: [1,3,2]

Follow up: Recursive solution is trivial, could you do it iteratively?

Accepted

558,021

Submissions

942,337

 

/**
 * Definition for a binary tree node.
 * 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;
        inorder(root,res);
        return res;
    }
    
    void inorder(TreeNode *root, vector<int> &res)
    {
        if(NULL==root)return;
        inorder(root->left,res);
        res.push_back(root->val);
        inorder(root->right,res);
    }
};

 

这可能是lc上最easy的medium了吧....