[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。

/**
 * 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 inorder(TreeNode *root, TreeNode *&pos1, TreeNode *&pos2, TreeNode *&pre) {
        if (root == NULL) return;
        inorder(root->left, pos1, pos2, pre);
        if (pre != NULL && pre->val > root->val) {
            if (pos1 == NULL) {
                pos1 = pre;
                pos2 = root;
            } else {
                pos2 = root;
            }
        }
        pre = root;
        inorder(root->right, pos1, pos2, pre);
    }
    
    void recoverTree(TreeNode *root) {
        TreeNode *pos1, *pos2, *pre;
        pos1 = pos2 = pre = NULL;
        inorder(root, pos1, pos2, pre);
        swap(pos1->val, pos2->val);
    }
};