## Monday, September 21, 2015

### LeetCode OJ - Convert Sorted Array to Binary Search Tree

Problem:

Solution:

I could have approached this problem using divide and conquer and it would have been much simpler. But I am not sure if I do that will the tree be strictly balanced.

Instead, I build a 'complete' binary tree and then by definition it has to be balanced. I used the heap trick to arrange the tree nodes in a array so that I can build the structure easily, and then I used the in order iterator trick to go through the list of nodes to put the values in.

Code:

#include "stdafx.h"

// https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree/

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

using namespace std;

namespace _LEET_CONVERT_SORTED_ARRAY_TO_BINARY_SEARCH_TREE
{
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution
{
public:
TreeNode* sortedArrayToBST(vector<int>& nums)
{
int size = nums.size();
if (size == 0)
{
return NULL;
}

/* allocate nodes and index them */
TreeNode** all_nodes = new TreeNode*[size];
for (int i = 0; i < size; i++)
{
all_nodes[i] = new TreeNode(-1);
}

all_nodes = all_nodes - 1;

/* build the tree structure */
TreeNode* root = all_nodes[1];
for (int i = 1; i <= size; i++)
{
int left = i * 2;
int right = left + 1;
if (left <= size)
{
all_nodes[i]->left = all_nodes[left];
}
else
{
all_nodes[i]->left = NULL;
}
if (right <= size)
{
all_nodes[i]->right = all_nodes[right];
}
else
{
all_nodes[i]->right = NULL;
}
}

/* build the tree values */
int t = 1;
int j = 0;
while (t * 2 <= size)
{
t = t * 2;
}

while (t != 0)
{
all_nodes[t]->val = nums[j++];
if (t * 2 + 1 <= size)
{
t = t * 2 + 1;
while (t * 2 <= size)
{
t = t * 2;
}
}
else
{
while (t % 2 == 1)
{
t = t / 2;
}
t = t / 2;
}
}

all_nodes = all_nodes + 1;
delete[] all_nodes;
return root;
}
};
};

using namespace _LEET_CONVERT_SORTED_ARRAY_TO_BINARY_SEARCH_TREE;

int LEET_CONVERT_SORTED_ARRAY_TO_BINARY_SEARCH_TREE()
{
int case1[] = { 1, 2, 3, 4, 5, 6, 7, 8 };

Solution s;
TreeNode* root = s.sortedArrayToBST(vector<int>(case1, case1 + _countof(case1)));
return 0;
}