Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
解题思路:此题意思就是给你一组排序好的数组转化为平衡二叉树,既然都是排列好的数组了,我们取中间的元素作为根结点,去中间元素的左边区域的中间元素作为左子结点,去中间元素的右边区域的中间元素作为右子结点。以此类推,使用递归,解之!!!
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: void convertToBST(TreeNode *&root,vector<int> &num,int pStart,int pEnd) { if(pStart>pEnd) return; int pMid=(pStart+pEnd)/2; root=new TreeNode(num[pMid]); convertToBST(root->left,num,pStart,pMid-1); convertToBST(root->right,num,pMid+1,pEnd); } TreeNode *sortedArrayToBST(vector<int> &num) { if(num.size()<=0) return NULL; int pStart=0; int pEnd=num.size()-1; TreeNode *root; convertToBST(root,num,pStart,pEnd); return root; } };