[Leetcode][Tree][Flatten Binary Tree to Linked List ]

按照前序遍历的顺序把树用right连起来。

本来想了半天，一点思路都没有，总觉得Inplace的解法一般都非常巧妙。

后来我突发灵感，决定用一个变量记录当前访问到哪个点，真是太机智了~~

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     void flatten(TreeNode *root) {
13         now = NULL;
14         findRes(root);
15     }
16 private:
17     TreeNode * now;
18     void findRes(TreeNode *root) {
19         if (root == NULL) {
20             return;
21         }
22         findRes(root->right);
23         findRes(root->left);
24         root->left = NULL;
25         root->right = now;
26         now = root;
27     }
28 };

一次AC之后看了题解，发现大家的思路差别好大！其中迭代的那个版本感觉很清晰。又写了一遍，同样是一次AC。

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     void flatten(TreeNode *root) {
13         if (root == NULL) {
14             return;
15         }
16         stack<TreeNode*> s;
17         s.push(root);
18         while (!s.empty()) {
19             TreeNode * now = s.top();
20             s.pop();
21             if (now->right != NULL) {
22                 s.push(now->right);
23             }
24             if (now->left != NULL) {
25                 s.push(now->left);
26             }
27             now->left = NULL;
28             if (!s.empty()) {
29                 now->right = s.top();
30             }
31         }
32     }
33 };

[Leetcode][Tree][Flatten Binary Tree to Linked List ]