【leetcode】108. Convert Sorted Array to Binary Search Tree
2016-08-18 16:51
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一、题目描述
Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
题目解读:有一个升序的数组,用这个数组构造一棵平衡的二叉搜索树BST。
思路:采用递归,因为是有序的,找到中值作为根节点,左边的为左子树,右边的为右子树。递归结束条件为如果数组为空,返回空指针。
c++代码(20ms,21.89%)
/**
* 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) {
return sortToBST(nums, 0, nums.size()-1);
}
TreeNode* sortToBST(vector<int>& nums,int first, int last){
if(first > last)
return NULL;
int mid=(first+last)/2;
TreeNode* node=new TreeNode(nums[mid]);
node->left=sortToBST(nums, first, mid-1);
node->right=sortToBST(nums, mid+1, last);
return node;
}
};
Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
题目解读:有一个升序的数组,用这个数组构造一棵平衡的二叉搜索树BST。
思路:采用递归,因为是有序的,找到中值作为根节点,左边的为左子树,右边的为右子树。递归结束条件为如果数组为空,返回空指针。
c++代码(20ms,21.89%)
/**
* 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) {
return sortToBST(nums, 0, nums.size()-1);
}
TreeNode* sortToBST(vector<int>& nums,int first, int last){
if(first > last)
return NULL;
int mid=(first+last)/2;
TreeNode* node=new TreeNode(nums[mid]);
node->left=sortToBST(nums, first, mid-1);
node->right=sortToBST(nums, mid+1, last);
return node;
}
};
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