19.3.2 [LeetCode 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?

题意

给出一棵标准BST树中两个结点互换后的树，要求互换前的BST树

题解

class Solution {
public:
    TreeNode*big = NULL, *small = NULL, *tmp;
    TreeNode*last=NULL;
    void search(TreeNode*root) {
        if (big&&small||root==NULL)return;
        if (root->left)
            search(root->left);
        if (last&&root->val < last->val) {
            if (big == NULL) {
                big = last;
                tmp = root;
            }
            else if (small == NULL) {
                small = root;
                return;
            }
        }
        last = root;
        if (root->right)
            search(root->right);
    }
    void recoverTree(TreeNode* root) {
        search(root);
        if (!small)
            small = tmp;
        swap(big->val, small->val);
        return;
    }
};

这题应该可以直接换值吧？如果不能的话就需要再改改了