## Sunday, November 6, 2016

### LeetCode OJ - Recover Binary Search Tree

Problem:

Analysis:

LeetCode claimed this is a hard problem, but it isn't so hard at all. All we need to do is to find the bad node that is too big, and the bad node that is too small, and then swap it.

Consider that in order traversal, if exactly one pair is swapped, then we should see this:

$\cdots, a, b, \cdots, c, d, \cdots$

$b < a$ and also $c > d$

In that case, we know we should swap a with d.

Solution:

To spot those pattern, all we need to do is to keep track of the previous node. Note that we have a special case that the node being exchanged are adjacent, so we are finding only one pair that is out of order, in that case, simply swapping that pair.

Code:

#include "stdafx.h"

// https://leetcode.com/problems/recover-binary-search-tree/

#include "LEET_RECOVER_BINARY_SEARCH_TREE.h"
#include <algorithm>
#include <iostream>
#include <sstream>
#include <vector>
#include <string>

using namespace std;

namespace _LEET_RECOVER_BINARY_SEARCH_TREE
{
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution {
private:
TreeNode* last;
public:
void recoverTree(TreeNode* root)
{
last = nullptr;
walk(root);
}
void walk(TreeNode* root) {
if (root == nullptr)
{
return;
}
else
{
walk(root->left);
if (last != nullptr)
{
if (root->val < last->val)
{
{
}
else
{
}
}
}
if (root != nullptr)
{
last = root;
}
walk(root->right);
}
}
};
};

using namespace _LEET_RECOVER_BINARY_SEARCH_TREE;

int LEET_RECOVER_BINARY_SEARCH_TREE()
{
Solution solution;
TreeNode a(4);
TreeNode b(2);
TreeNode c(6);
TreeNode d(7);
TreeNode e(3);
TreeNode f(5);
TreeNode g(1);
a.left = &b;
a.right = &c;
b.left = &d;
b.right = &e;
c.left = &f;
c.right = &g;
solution.recoverTree(&a);
cout << a.val;
cout << b.val;
cout << c.val;
cout << d.val;
cout << e.val;
cout << f.val;
cout << g.val;
return 0;
}