99. Recover Binary Search Tree

题目

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?


confused what"{1,#,2,3}"means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / \
 2   3
    /
   4
    \
     5

The above binary tree is serialized as"{1,2,3,#,#,4,#,#,5}". 

解析

• 一种非递归且不使用栈，空间复杂度为O(1)的遍历方法
• [LeetCode] Recover Binary Search Tree 复原二叉搜索树

class Solution_99 {
public:
    void recoverTree(TreeNode* root) {

        TreeNode* first = NULL, *second = NULL, *parenet = NULL;
        TreeNode*cur, *pre; //中序遍历
        cur = root;

        while (cur)
        {
            if (cur->left==NULL)
            {
                if (parenet&&parenet->val>cur->val)
                {
                    if (!first)
                    {
                        first = parenet;
                    }
                    second = cur;
                }
                parenet = cur;
                cur = cur->right;
            }
            else
            {
                pre = cur->left; //找到左子树的最右节点
                while (pre->right&&pre->right!=cur)
                {
                    pre = pre->right;
                }

                if (!pre->right)
                {
                    pre->right = cur;
                    cur = cur->left;
                }
                else
                {
                    pre->right = NULL; //恢复树结构
                    if (parenet&&parenet->val > cur->val)
                    {
                        if (!first)
                        {
                            first = parenet;
                        }
                        second = cur;
                    }
                    parenet = cur;
                    cur = cur->right;
                }
            }
        }
        if (first&&second)
        {
            swap(first->val, second->val);
        }
    }
};

题目来源

• 99. Recover Binary Search Tree