/* 题意:把一个有序的数组转化成平衡二叉树 解法:递归 */ /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: TreeNode *dfs(vector<int> &num,int left,int right){ int index = left+(right-left)/2; TreeNode *node = new TreeNode(num[index]); if(left == right){ return node; } if(left<=index-1) node->left = dfs(num,left,index-1); if(index+1<=right) node->right = dfs(num,index+1,right); return node; } TreeNode *sortedArrayToBST(vector<int> &num) { if(!num.size()) return NULL; return dfs(num,0,num.size()-1); } };