[Leetcode] Recover Binary Search Tree

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?

 

中序遍历，对于BST，其中序序列是有序的，所以只要找出中序序列中的逆序对，但是要注意保存中间节点，可以使用全局变量，也可以使用引用。

 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 inorder(TreeNode *root, TreeNode *&pos1, TreeNode *&pos2, TreeNode *&pre) {

13         if (root == NULL) return;

14         inorder(root->left, pos1, pos2, pre);

15         if (pre != NULL && pre->val > root->val) {

16             if (pos1 == NULL) {

17                 pos1 = pre;

18                 pos2 = root;

19             } else {

20                 pos2 = root;

21             }

22         }

23         pre = root;

24         inorder(root->right, pos1, pos2, pre);

25     }

26     

27     void recoverTree(TreeNode *root) {

28         TreeNode *pos1, *pos2, *pre;

29         pos1 = pos2 = pre = NULL;

30         inorder(root, pos1, pos2, pre);

31         swap(pos1->val, pos2->val);

32     }

33 };