Binary Tree Preorder Traversal

问题描述：

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

For example:
Given binary tree {1,#,2,3},

   1    \     2    /   3

return [1,2,3].

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

C++:非递归实现，不过超时了。

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    vector<int> preorderTraversal(TreeNode *root) {        vector<int> pretr;        if(root == NULL)            return pretr;        stack<TreeNode *> node;        node.push(root);        while(!node.empty()){            TreeNode *p = node.top();            pretr.push_back(p->val);            node.pop();            if(p->right) node.push(p);            if(p->left) node.push(p);        }        return pretr;    }};

C++递归实现

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    vector<int> preorderTraversal(TreeNode *root) {        vector<int> pretr;        process(root,pretr);        return pretr;            }    void process(TreeNode *root,vector<int> &pretr){        if(root == NULL) return;        TreeNode *p;        p = root;        pretr.push_back(p->val);        process(p->left,pretr);        process(p->right,pretr);    }};

python:

# Definition for a  binary tree node# class TreeNode:#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Solution:    # @param root, a tree node    # @return a list of integers    def preorderTraversal(self, root):        pretr = []        self.process(root,pretr)        return pretr        def process(self,root,ls):        if(root == None):            return;        ls.append(root.val)        self.process(root.left,ls)        self.process(root.right,ls)