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

 1 class Solution {
 2 public:
 3     TreeNode*big = NULL, *small = NULL, *tmp;
 4     TreeNode*last=NULL;
 5     void search(TreeNode*root) {
 6         if (big&&small||root==NULL)return;
 7         if (root->left)
 8             search(root->left);
 9         if (last&&root->val < last->val) {
10             if (big == NULL) {
11                 big = last;
12                 tmp = root;
13             }
14             else if (small == NULL) {
15                 small = root;
16                 return;
17             }
18         }
19         last = root;
20         if (root->right)
21             search(root->right);
22     }
23     void recoverTree(TreeNode* root) {
24         search(root);
25         if (!small)
26             small = tmp;
27         swap(big->val, small->val);
28         return;
29     }
30 };

这题应该可以直接换值吧？如果不能的话就需要再改改了

标签：Binary,Search,NULL,19.3,big,search,small,null,root
来源： https://www.cnblogs.com/yalphait/p/10460729.html