LeetCode [99] Recover Binary Search Tree

 99. Recover Binary Search Tree

Hard

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?

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/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */ class Solution { public: void recoverTree(TreeNode* root) { TreeNode *e1 = NULL, *e2 = NULL, *prev = NULL; int lb = INT_MIN; dfs(root, prev, e1, e2, lb); swap(e1->val, e2->val); } void dfs(TreeNode* root, TreeNode *&prev, TreeNode *&e1, TreeNode *&e2, int &lb){ if(!root) return; dfs(root->left, prev, e1, e2, lb); if(root->val<lb){ if(!e1){ e1 = prev; } e2 = root; } lb = max(root->val, lb); prev = root; dfs(root->right, prev, e1, e2, lb); } };