[LeetCode] Binary Tree Inorder Traversal [递归版]

题目：

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}".

解答：

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {private:    vector<int> result;public:    vector<int> inorderTraversal(TreeNode *root) {        return tmpFuction(root, result);    }        vector<int> tmpFuction(TreeNode *root, vector<int> &result) {        if(root == NULL) {            return result;        }        tmpFuction(root->left, result);        result.push_back(root->val);        tmpFuction(root->right, result);        return result;    }};

如果用递归来做就没什么好说的了，其程序编写方法与Binary Tree Preorder Traversal一致