Leetcode 144. Binary Tree Preorder Traversal

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?

s思路：
1. pre-order之前有遇到过：先根，再左，后右。recursive确实容易。iterative稍微比in-order麻烦一些，需要用stack，关键是使用两个指针pnow和pre，防止做回头路！
2. 这个pre和pnow的关系是：pnow访问了左边，移动到右边之前，pre被赋值为pnow；当pnow访问了右边，移动到上层之前，pre被赋值为pnow。

//方法1:recursive: pre-orderclass Solution {public:    void helper(vector<int>&res,TreeNode* root){        if(!root) return;        res.push_back(root->val);//root        helper(res,root->left);//left        helper(res,root->right);//right         }    vector<int> preorderTraversal(TreeNode* root) {        //        vector<int> res;        helper(res,root);        return res;    }};//方法2:iterative: pre-order;stack;pre+pnowclass Solution {public:    vector<int> preorderTraversal(TreeNode* root) {        //        vector<int> res;        stack<TreeNode*> ss;        TreeNode* pnow=root,*pre=NULL;        while(!ss.empty()||pnow){            while(pnow){                res.push_back(pnow->val);                ss.push(pnow);                pnow=pnow->left;                }               pnow=ss.top();            if(pnow->right&&pnow->right!=pre){                pnow=pnow->right;               }else{                pre=pnow;                ss.pop();                pnow=NULL;              }        }               return res;    }};