108. Convert Sorted Array to Binary Search Tree

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

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升序数组a[],构造平衡二叉树,左右子树高度差不超过1.

思路:

利用先序构建二叉树的方法,将数组的中间节点a[mid]作为根节点root的值,

root的左子树由a[0,...mid-1]构造

root的右子树由a[mid+1,...a.size-1]构造

最后返回root

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/**
 * 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 nullptr;
        TreeNode *root = help_saTobst(nums,0,nums.size()-1);
        return root;
    }
    TreeNode* help_saTobst(vector<int> &nums,int s,int end){
        if(s>end) return nullptr;
        int mid = (s+end)/2;
        TreeNode *root = new TreeNode(nums[mid]);
        root->left = help_saTobst(nums,s,mid-1);
        root->right = help_saTobst(nums,mid+1,end);
        return root;
    }
};