94. Binary Tree Inorder Traversal 解题记录

题目描述：

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

For example:
Given binary tree [1,null,2,3],

   1
    \
     2
    /
   3

return [1,3,2].

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

解题思路：

利用递归一路从左往下，然后添加当前节点，最后再从右往下，遇到节点为空则返回。

代码：

/**
 * 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:
    void inorder(vector<int> &nums, TreeNode* root){
        if(root){
            inorder(nums, root->left);  //一路左
            nums.push_back(root->val);  //添加当前节点
            inorder(nums, root->right);  //最后往右
        }
    }
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> nums;
        inorder(nums, root);
        return nums;
    }
};