108. Convert Sorted Array to Binary Search Tree

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108. Convert Sorted Array to Binary Search Tree

题目

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

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example:

Given the sorted array: [-10,-3,0,5,9],

One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

      0
     / \
   -3   9
   /   /
 -10  5

解析

- vector的初始化方法，递归创建左右子树的思想！

// Convert Sorted Array to Binary Search Tree
class Solution_108 {
public:
	TreeNode* sortedArrayToBST(vector<int>& nums) {
		if (nums.size()==0)
		{
			return NULL;
		}
		if (nums.size()==1)
		{
			TreeNode* temp = new TreeNode(nums[0]);
			return temp;
		}

		int mid = nums.size() / 2;

		TreeNode* root = new TreeNode(nums[mid]);

		auto leftTree = vector<int>(nums.begin(), nums.begin() + mid);//最后一个迭代器指向最后一个元素的下一个位置
		auto rightTree = vector<int>(nums.begin() + mid + 1, nums.end());

		root->left = sortedArrayToBST(leftTree);
		root->right = sortedArrayToBST(rightTree);
		
		return root;
	}
};