leetcode Convert Sorted Array to Binary Search Tree-优快云博客

本文详细介绍了如何使用二分法将一个已排序的整数数组转换为一个高度平衡的二叉搜索树（BST）。通过每次选择中间元素作为根节点，确保最终构造出的BST高度平衡。

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Convert Sorted Array to Binary Search Tree

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

本题是二分法的高级应用，每次选择root都是从数列的中间选择，那么最后构造出来的二叉树一定是高度平衡的BST了。

/**
 * 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) 
	{
		return biSortedArrayToBST(num, 0, num.size()-1);
	}

	TreeNode *biSortedArrayToBST(vector<int> &num, int low, int up)
	{
		if (low > up) return NULL;
		int mid = low + ((up-low)>>1);
		TreeNode *r = new TreeNode(num[mid]);

		r->left = biSortedArrayToBST(num, low, mid-1);
		r->right = biSortedArrayToBST(num, mid+1, up);

		return r;
	}
};

//2014-2-16 update
	TreeNode *sortedArrayToBST(vector<int> &num) 
	{
		return toBST(num, 0, num.size()-1);//下标num.size()-1
	}
	TreeNode *toBST(vector<int> &num, int n1, int n2)
	{
		if (n1 > n2) return nullptr;
		int mid = n1 + ((n2-n1)>>1);
		TreeNode *root = new TreeNode(num[mid]);

		root->left = toBST(num, n1, mid-1);
		root->right = toBST(num, mid+1, n2);
		return root;
	}