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?

分析：二叉树的遍历算法可用的场景很多，此题是个很好的例子。BST的一个特点是它的中序遍历是ascending的，所以我们可以通过BST的中序比遍历确定是哪两个元素的值互换，这里要求是constant space，所以很自然想到O(1)空间复杂度的Morris中序遍历。我们用一个pair记录值互换的两个元素，最后交换他们的值即可。代码如下：

class Solution {
public:
    void recoverTree(TreeNode *root) {
        TreeNode *cur = root;
        TreeNode *prev = NULL;
        pair<TreeNode *, TreeNode *> broken;
        
        while(cur){
            if(cur->left == NULL){
                detect(broken, prev, cur);
                prev = cur;
                cur = cur->right;
            }else{
                TreeNode *node = cur->left;
                for(; node->right && node->right != cur; node = node->right);
                if(node->right == NULL){
                    node->right = cur;
                    cur = cur->left;
                }else{
                    node->right = NULL;
                    detect(broken, prev, cur);
                    prev = cur;
                    cur = cur->right;
                }
            }
        }
        swap(broken.first->val, broken.second->val);
    }
    
    void detect(pair<TreeNode *, TreeNode *> &broken, TreeNode *prev, TreeNode *cur){
        if(prev != NULL && prev->val > cur->val){
            if(broken.first == NULL){
                broken.first = prev;
            }
            broken.second = cur;
        }
    }
};