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?
Solution: Inorder traversal
Using inorder traversal to find two nodes that have val < prev.val
Time complexity: O(n)
Space complexity: O(h)
C++
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// Author: Huahua class Solution { public: void recoverTree(TreeNode *root) { inorder(root); swap(first->val, second->val); } void inorder(TreeNode* root) { if (!root) return; inorder(root->left); if (prev && prev->val > root->val) { if (!first) first = prev; second = root; } prev = root; inorder(root->right); } private: TreeNode* first; TreeNode* second; TreeNode* prev; }; |
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