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....