Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
有序数组变二叉平衡搜索树，不难，递归就行。每次先序建立根节点（取最中间的数），然后用子区间划分左右子树。
一次就AC了
注意：new 结构体的时候对于
struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {} };
要用 TreeNode * root = new TreeNode(123); 与构造函数的形式要一致。
#include <iostream> #include <vector> #include <algorithm> #include <queue> #include <stack> using namespace std; //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 *sortedArrayToBST(vector<int> &num) { if(num.empty()) { return NULL; } int nr = num.size()/2; TreeNode * root = new TreeNode(num[nr]); vector<int> numl(num.begin(), num.begin() + nr); vector<int> numr(num.begin() + nr + 1, num.end()); root->left = sortedArrayToBST(numl); root->right = sortedArrayToBST(numr); return root; } }; int main() { Solution s; vector<int> num; num.push_back(1); num.push_back(2); num.push_back(3); num.push_back(4); num.push_back(5); num.push_back(6); num.push_back(7); TreeNode * ans = s.sortedArrayToBST(num); return 0; }
【leetcode】Convert Sorted Array to Binary Search Tree (easy)
原文：http://www.cnblogs.com/dplearning/p/4127261.html