Convert Sorted Array to Binary Search Tree - leetcode

Convert Sorted Array to Binary Search Tree

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Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

思路： 利用数组有序，然后二分建立二叉查找树

#include <iostream>
#include <vector>

using namespace std;

//Definition for binary 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> &num) {
// Note: The Solution object is instantiated only once and is reused by each test case.
if (num.empty ())
return NULL;
return createTree (num, 0, num.size()-1);
}
TreeNode * createTree (const vector <int> & num, int lhs, int rhs ) {
if (lhs > rhs )
return NULL;
int mid = lhs + ((rhs - lhs) >> 1);
TreeNode * node = new TreeNode (num.at(mid));
node->left = createTree (num, lhs, mid - 1);
node->right = createTree (num, mid + 1, rhs);
return node;
}
void delTree (TreeNode * tree) {
if (tree == NULL)
return ;
delTree (tree->left);
delTree (tree->right);
cout << "delete : " << tree->val << endl;
delete tree;
tree = NULL;
}
void inOrder (TreeNode * p) {
if (p) {
inOrder (p->left);
cout << "traverse : " << p->val << endl;;
inOrder (p->right);
}
}
};

int main () {
Solution s;
int  a[] = {1, 2, 3, 4, 5, 6};

vector<int> v (a, a + sizeof(a) / sizeof (a[0]));

TreeNode * tree = s.sortedArrayToBST (v);

s.inOrder (tree);

s.delTree (tree);

return 0;
}