108ConvertSortedArraytoBinarySearchTree.org

[#C] 108 Convert Sorted Array to Binary Search Tree - LeetCode

题目

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

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example:

Given the sorted array: [-10,-3,0,5,9],

One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

      0
     / \
   -3   9
   /   /
 -10  5

思路

对于二分搜索树来说，它的中序遍历输出就是一个排好序的数组，所以在数组的中间就是根节点，左边部分是左子树，右边部分是右子树。通过递归就能实现把有序数组转换为二分搜索树。剩下的左边部分的中间就是左节点，同理剩下右边部分的中间恰好是右节点。

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        return helper(nums, 0, nums.length - 1);
    }
 
    public TreeNode helper(int[] nums, int l, int r) {
        if (l > r) return null;
        int mid = l + (r - l) / 2;
        TreeNode node = new TreeNode(nums[mid]);
        node.left = helper(nums, l, mid - 1);
        node.right = helper(nums, mid + 1, r);
        return node;
    }
}

CPP

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        return helper(nums, 0, nums.size() - 1);
    }
 
    TreeNode* helper(vector<int>& nums, int l, int r) {
        if (l > r) return NULL;
        int mid = l + (r - l) / 2;
        TreeNode* node = new TreeNode(nums[mid]);
        node->left = helper(nums, l, mid - 1);
        node->right = helper(nums, mid + 1, r);
        return node;
    }
};