[LeetCode] Convert Sorted Array to Binary Search Tree

原创于 2013-10-18 11:15:52 发布 · 320 阅读

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本文介绍如何将一个升序排列的数组转换为高度平衡的二叉搜索树。通过递归方式，选取数组中间元素作为根节点，并将左右两部分数组分别构建为左子树和右子树。

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

从有序数组构造二叉搜索树

二叉搜索树要是平衡的，因此每次以数组中间元素作为跟，然后数组左半部分元素作为左子树递归，右半部分作为右子树递归

递归即简洁

/**
 * 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) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        return build(num,0,num.size()-1);
    }
    
    TreeNode *build(vector<int> &num, int start, int end) {
        if(start > end) return NULL;
        if(start == end) return new TreeNode(num[start]);
        
        int middle = start+(end-start)/2;
        TreeNode *root = new TreeNode(num[middle]);
        root->left = build(num,start,middle-1);
        root->right = build(num,middle+1,end);
    }
};