[LeetCode]Recover Binary Search Tree

题目描述:(链接)

Two elements of a binary search tree (BST) are swapped by mistake.

Recover the tree without changing its structure.

Note:
A solution using O(n) space is pretty straight forward. Could you devise a constant space solution?

解题思路:

空间复杂度是O(n), 利用中序遍历，将节点保存在一个数组中，然后遍历数组，找出错位的两个节点，交换值。

 1 /**
 2  * Definition for a binary tree node.
 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 recoverTree(TreeNode* root) {
13         vector<TreeNode *> cache1;
14         stack<TreeNode *> cache2;
15         TreeNode *p = root;
16         
17         while (!cache2.empty() || p != nullptr) {
18             if (p != nullptr) {
19                 cache2.push(p);
20                 p = p->left;
21             } else {
22                 p = cache2.top();
23                 cache2.pop();
24                 cache1.push_back(p);
25                 p = p->right;
26             }
27         }
28         
29         int len = cache1.size();
30         TreeNode *first;
31         for (int i = 0; i < len - 1; ++i) {
32             TreeNode *tmp1 = cache1[i];
33             TreeNode *tmp2 = cache1[i + 1];
34             if (tmp1->val > tmp2->val) {
35                 first = tmp1;
36                 break;
37             }
38         }
39         
40         TreeNode *second;
41         for (int i = len - 1; i > 0; --i) {
42             TreeNode *tmp1 = cache1[i];
43             TreeNode *tmp2 = cache1[i - 1];
44             if (tmp1->val < tmp2->val) {
45                 second = tmp1;
46                 break;
47             }
48         }
49         
50         if (first != nullptr && second != nullptr) {
51             swap(first->val, second->val);
52         }
53     }
54 };