LeetCode – Refresh – Convert Sorted Array to Binary Search Tree

转载于 2015-03-19 05:38:00 发布 · 50 阅读

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原文链接：http://www.cnblogs.com/shuashuashua/p/4349278.html

本文介绍了一种使用二分查找的方法将有序数组转换为平衡二叉搜索树的算法。通过递归地找到中间元素作为根节点，并将左右两侧的子数组分别构建成左子树和右子树。

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It is kind of binary search. Since this is a sorted array, we can treat it as inorder traversal. And we can define the root node for the Tree.

So find the middle element as the root, then keep doing recursion.

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     TreeNode *getTree(const vector<int> &num, int start, int end) {
13         if (start > end) return NULL;
14         int mid = (start + end)/2;
15         TreeNode *root = new TreeNode(num[mid]);
16         root->left = getTree(num, start, mid-1);
17         root->right = getTree(num, mid+1, end);
18         return root;
19     }
20     TreeNode *sortedArrayToBST(vector<int> &num) {
21         return getTree(num, 0, num.size()-1);
22     }
23 };