LeetCode 99: Recover Binary Search Tree

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?

解题思路

利用二叉搜索树中序遍历的结果是一个有序递增序列的性质，在中序遍历的过程中寻找需要交换数据的两个结点。在这过程中有两种情况需要考虑：需要交换的元素在中序下相邻，如

[1, 2, 4, 3, 5]

；需要交换的元素在中序下不相邻，如

[1, 5, 3, 4, 2

。

代码如下：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
private:
    void inOrderFindErrorNode(TreeNode *root, TreeNode*& e1, 
                                TreeNode*& e2, TreeNode*& pre) {
        if (root == NULL) return;

        inOrderFindErrorNode(root->left, e1, e2, pre);

        if (pre != NULL && pre->val >= root->val) {
            e2 = root;
            if (e1 == NULL) {
                e1 = pre;
            }
        }

        pre = root;
        inOrderFindErrorNode(root->right, e1, e2, pre);
    }

public:
    void recoverTree(TreeNode* root) {
       TreeNode *pre = NULL, *e1 = NULL, *e2 = NULL;
       inOrderFindErrorNode(root, e1, e2, pre);
       swap(e1->val, e2->val);
    }
};