【ATT】Recover Binary Search Tree

Q: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?

A: 如果交换了，inorder seq中的逆序对可能是1对(父子节点交换)，或者2对。

    void recoverTree(TreeNode *root) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        TreeNode* n1 = NULL,*n2 = NULL,*prev = NULL;
        getErrorNodes(root,n1,n2,prev);
        if(n1&&n2)
        {
            int tmp = n1->val;
            n1->val = n2->val;
            n2->val =  tmp;
        }
    }
    
    void getErrorNodes(TreeNode* root,TreeNode* &n1,TreeNode* &n2,TreeNode*& prev)
    {
        if(!root)
            return;
        getErrorNodes(root->left,n1,n2,prev); //prev是inoder seq的前一个节点
        if(prev&&prev->val>root->val) //发现逆序对。
        {
            n2 = root;
            if(!n1)
                n1 = prev;
        }
        prev = root;
        getErrorNodes(root->right,n1,n2,prev);
    }

　　

如果要求O(1)的空间复杂度，把inorder recursive改为递归做，即可。