Leetcode 108 Convert Sorted Array to Binary Search Tree
2016-10-12 17:03
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Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
将有序数组转化为二叉搜索树,其实题目说的不清楚,应该是二叉平衡树。
二分的思想。
/**
* 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* dfs(vector<int>& nums,int l,int r)
{
if(l>r) return NULL;
int mid=(l+r)>>1;
TreeNode* p=new TreeNode(nums[mid]);
p->left=dfs(nums,l,mid-1);
p->right=dfs(nums,mid+1,r);
return p;
}
TreeNode* sortedArrayToBST(vector<int>& nums) {
return dfs(nums,0,nums.size()-1);
}
};
将有序数组转化为二叉搜索树,其实题目说的不清楚,应该是二叉平衡树。
二分的思想。
/**
* 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* dfs(vector<int>& nums,int l,int r)
{
if(l>r) return NULL;
int mid=(l+r)>>1;
TreeNode* p=new TreeNode(nums[mid]);
p->left=dfs(nums,l,mid-1);
p->right=dfs(nums,mid+1,r);
return p;
}
TreeNode* sortedArrayToBST(vector<int>& nums) {
return dfs(nums,0,nums.size()-1);
}
};
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