leetcode94. Binary Tree Inorder Traversal

題目描述：

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]

.

分析：遞歸求解。。。

c++代碼：

/**
 * 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 {
private:
vector<int> result;
public:
    vector<int> inorderTraversal(TreeNode* root) {
        result.clear();
        inorder(root);
        return result;
    }
    void inorder(TreeNode* root)
    {
        if(root==NULL) return;
        inorder(root->left);
        result.push_back(root->val);
        inorder(root->right);
    }
};           

python代碼：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        self.result=[]
        self.inorder(root)
        return self.result
    def inorder(self,root):
        if root==None:
            return 
        self.inorder(root.left)
        self.result.append(root.val)
        self.inorder(root.right)           

（完）