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Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
思路比较简单，就是每次以中位数为根节点，然后左边即左子树，右边是右子树，递归下去即可
超时代码（不知道为啥）：
class Solution{ public: void dfs(TreeNode* &root,vector<int> num,int start,int end){ int mid=(start+end)/2; root=new TreeNode(num[mid]); if(start<=mid-1) dfs(root->left,num,start,mid-1); if(mid+1<=end) dfs(root->right,num,mid+1,end); return; } TreeNode *sortedArrayToBST(vector<int> &num) { if(num.empty()) return NULL; TreeNode* root=NULL; dfs(root,num,0,num.size()-1); return root; } };
AC代码：
class Solution{ public: TreeNode *createTree(vector<int> &num, int start, int end) { // if (start > end) return NULL; int mid = (start+end)/2; TreeNode *node = new TreeNode(num[mid]); if(start<=mid-1) node->left = createTree(num, start, mid-1); if(mid+1<=end) node->right = createTree(num, mid+1, end); return node; } TreeNode *sortedArrayToBST(vector<int> &num) { if(num.empty()) return NULL; return createTree(num,0,num.size()-1); } };
Convert Sorted Array to Binary Search Tree
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原文地址：http://www.cnblogs.com/fightformylife/p/4244783.html
