【Binary Tree Preorder Traversal】cpp

题目：

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?

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:
    vector<int> preorderTraversal(TreeNode* root) {
            vector<int> ret_l, ret_r;
            if ( root ){
                ret_l = Solution::preorderTraversal(root->left);
                ret_l.insert(ret_l.begin(), root->val);
                ret_r = Solution::preorderTraversal(root->right);
                ret_l.insert(ret_l.end(), ret_r.begin(), ret_r.end());
            }
            return ret_l;
    }
};

tips：

trivial的递归版

===================================

再来一个non trivial的迭代版

/**
 * 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:
    vector<int> preorderTraversal(TreeNode* root) {
            vector<int> ret;
            if ( !root ) return ret;
            stack<TreeNode*> sta;
            sta.push(root);
            while ( !sta.empty() )
            {
                TreeNode *tmp = sta.top();
                sta.pop();
                ret.push_back(tmp->val);
                if (tmp->right) sta.push(tmp->right);
                if (tmp->left) sta.push(tmp->left);
            }
            return ret;
    }
};

tips:

把递归调用改成人工堆栈：

1. 把根节点压进栈

2. 栈定元素出栈，把val加入到ret中

3. 注意：先压right入栈，再压left入栈（保证压入栈的元素都是非NULL）

循环1~3，直到栈空。

==============================================

还有更高端一些的Morris遍历算法：特点是空间复杂度是O(1)

http://www.cnblogs.com/AnnieKim/archive/2013/06/15/morristraversal.html

=================================================

第二次过这道题，只写了一个递归的。

/**
 * 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:
        vector<int> preorderTraversal(TreeNode* root)
        {
            vector<int> ret;
            Solution::traversal(ret, root);
            return ret;
        }
        static void traversal(vector<int>& ret, TreeNode* root)
        {
            if (!root) return;
            ret.push_back(root->val);
            Solution::traversal(ret, root->left);
            Solution::traversal(ret, root->right);
        }
};