LeetCode: Recover Binary Search Tree

这题想了很久，看了网上答案，这段代码精髓在于findpos第一个pos是在他下一个node函数里才被发现的，因为第一个pos->val是变大了，所以p是pre，而第二个pos是在他自己本身的函数里被发现的，因为第二个pos->val是变小了，所以就是root。

 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 findpos(TreeNode *root, TreeNode * &pre, TreeNode * &p, TreeNode * &q) {
13         if (!root) return;
14         findpos(root->left, pre, p, q);
15         if (pre && pre->val > root->val) {
16             if (!p) p = pre;
17             q = root;
18         }
19         pre = root;
20         findpos(root->right, pre, p, q);
21     }
22     void recoverTree(TreeNode *root) {
23         // Start typing your C/C++ solution below
24         // DO NOT write int main() function
25         TreeNode *p, *q, *pre;
26         pre = p = q = NULL;
27         findpos(root, pre, p, q);
28         int tmp = p->val;
29         p->val = q->val;
30         q->val = tmp;
31     }
32 };

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     public int val;
 5  *     public TreeNode left;
 6  *     public TreeNode right;
 7  *     public TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public void RecoverTree(TreeNode root) {
12         TreeNode p = null, q = null, pre = null;
13         findpos(root, ref pre, ref p, ref q);
14         int tmp = p.val;
15         p.val = q.val;
16         q.val = tmp;
17     }
18     public void findpos(TreeNode root, ref TreeNode pre, ref TreeNode p, ref TreeNode q) {
19         if (root == null) return;
20         findpos(root.left, ref pre, ref p, ref q);
21         if (pre != null && pre.val > root.val) {
22             if (p == null) p = pre;
23             q = root;
24         }
25         pre = root;
26         findpos(root.right, ref pre, ref p, ref q);
27     }
28 }

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