99. Recover Binary Search Tree

• Difficulty: Hard

• Topics: Tree, Depth-first Search

• Similar Questions:

Problem:

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?

Solutions:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void recoverTree(TreeNode* root) {
        vector<TreeNode*> nodes;
        inOrder(root, nodes);
        TreeNode* first = nullptr;
        TreeNode* second = nullptr;
        for (int i = 1; i < nodes.size(); ++i) {
            if (nodes[i]->val <= nodes[i-1]->val) {
                if (first == nullptr) {
                    first = nodes[i-1];
                    second = nodes[i];
                } else {
                    second = nodes[i];
                }
            }
        }


        swap(first->val, second->val);


    }

private:
    void inOrder(TreeNode* root, vector<TreeNode*>& nodes) {
        if (root == nullptr)    return;
        inOrder(root->left, nodes);
        nodes.push_back(root);
        inOrder(root->right, nodes);
    }

};