Convert Sorted Array to Binary Search Tree
2017-03-19 17:15
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Description
Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
解题思路:
给定一个元素递增的数组,构造一棵平衡的二叉搜索树。既然数组已经有序,而且是平衡的二叉搜索树,所以我想到的便是一开始取数组的中间元素作为根节点,用中间元素左边的元素构造左子树,中间元素右边的元素构造右子树,这样就使根节点平衡并且可以是BST,接下来对左子树做同样的操作,选择左边元素的中间元素作为左子树的根节点,中间元素左边的元素构造左子树根节点的左子树,中间元素右边的元素构造右子树,到这里可以看出分治法的思想。程序代码如下:
Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
解题思路:
给定一个元素递增的数组,构造一棵平衡的二叉搜索树。既然数组已经有序,而且是平衡的二叉搜索树,所以我想到的便是一开始取数组的中间元素作为根节点,用中间元素左边的元素构造左子树,中间元素右边的元素构造右子树,这样就使根节点平衡并且可以是BST,接下来对左子树做同样的操作,选择左边元素的中间元素作为左子树的根节点,中间元素左边的元素构造左子树根节点的左子树,中间元素右边的元素构造右子树,到这里可以看出分治法的思想。程序代码如下:
/** * Definition for a binary tree node. * 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; if (size == 1) { TreeNode* leaf = new TreeNode(nums[0]); return leaf; } int mid = size / 2; vector<int> l_nums; //保存中间元素左边的元素 vector<int> r_nums; //保存中间元素右边的元素 for (int i = 0; i < mid; i++) { l_nums.push_back(nums[i]); } for (int i = mid + 1; i < size; i++) { r_nums.push_back(nums[i]); } TreeNode* l_tree = sortedArrayToBST(l_nums); //构造符合条件的左子树 TreeNode* r_tree = sortedArrayToBST(r_nums); //构造符合条件的右子树 TreeNode* root = new TreeNode(nums[mid]); root -> left = l_tree; root -> right = r_tree; return root; } };
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