Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
把有序数组转化成平衡的BST
找到数组中间的元素,作为根节点,则根节点左边是左子树,根节点右边是右子树,接着递归
/**
* 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* sortedArrayToBST(vector<int>& nums) {
if(nums.empty())return NULL;
return BuildTree(nums,0,nums.size()-1);
}
TreeNode* BuildTree(vector<int>& nums,int first,int end)
{
if(first>end)return NULL;
if(first==end)return new TreeNode(nums[first]);
int mid=(first+end)/2;
TreeNode* root=new TreeNode(nums[mid]);
root->left=BuildTree(nums,first,mid-1);
root->right=BuildTree(nums,mid+1,end);
return root;
}
};
leetcode No108. Convert Sorted Array to Binary Search Tree
原文:http://blog.csdn.net/u011391629/article/details/52296858