[WXM] LeetCode 99. Recover Binary Search Tree C++

99. Recover Binary Search Tree

Two elements of a binary search tree (BST) are swapped by mistake.

Recover the tree without changing its structure.

Example 1:

Input: [1,3,null,null,2]

   1
  /
 3
  \
   2

Output: [3,1,null,null,2]

   3
  /
 1
  \
   2

Example 2:

Input: [3,1,4,null,null,2]

  3
 / \
1   4
   /
  2

Output: [2,1,4,null,null,3]

  2
 / \
1   4
   /
  3

Follow up:

• A solution using O(n) space is pretty straight forward.
• Could you devise a constant space solution?

Approach

1. 题目大意二叉搜索树中有两个元素位置错误，需要你将它们重新交换回去，并不破坏原先的结构。这道题不难，二叉搜索数有一个性质就是中序即升序，那么你只要遍历升序的时候找到两个异常的元素结点即可。

Code

class Solution {
public:
    TreeNode* lastNode;
    TreeNode* firstNode;
    TreeNode* secondNode;
    void inorderTraval(TreeNode* root) {
        if (root == nullptr)return;
        inorderTraval(root->left);
        if (lastNode != nullptr&&firstNode == nullptr&&lastNode->val >= root->val)
            firstNode = lastNode;
        if (lastNode != nullptr&&firstNode != nullptr&&lastNode->val >= root->val)
            secondNode = root;
        lastNode = root;
        inorderTraval(root->right);
    }
    void recoverTree(TreeNode* root) {
        inorderTraval(root);
        swap(firstNode->val, secondNode->val);
    }
};