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。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void findpos(TreeNode *root, TreeNode * &pre, TreeNode * &p, TreeNode * &q) {
        if (!root) return;
        findpos(root->left, pre, p, q);
        if (pre && pre->val > root->val) {
            if (!p) p = pre;
            q = root;
        }
        pre = root;
        findpos(root->right, pre, p, q);
    }
    void recoverTree(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        TreeNode *p, *q, *pre;
        pre = p = q = NULL;
        findpos(root, pre, p, q);
        int tmp = p->val;
        p->val = q->val;
        q->val = tmp;
    }
};

 C#

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void RecoverTree(TreeNode root) {
        TreeNode p = null, q = null, pre = null;
        findpos(root, ref pre, ref p, ref q);
        int tmp = p.val;
        p.val = q.val;
        q.val = tmp;
    }
    public void findpos(TreeNode root, ref TreeNode pre, ref TreeNode p, ref TreeNode q) {
        if (root == null) return;
        findpos(root.left, ref pre, ref p, ref q);
        if (pre != null && pre.val > root.val) {
            if (p == null) p = pre;
            q = root;
        }
        pre = root;
        findpos(root.right, ref pre, ref p, ref q);
    }
}