108. Convert Sorted Array to Binary Search Tree / 109. Convert Sorted List to Binary Search Tree

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.

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

My Solutions:

找到sorted array中点作为root，root.left是root左边的数，root.left 是root右边的数

class Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        if (nums == null) return null;
        return helper(nums, 0, nums.length - 1);
    }
    
    private TreeNode helper(int[] nums, int start, int end) {
        if (start > end) return null;
        TreeNode node = new TreeNode(nums[(start + end) / 2]);
        node.left = helper(nums, start, (start + end) / 2 - 1);
        node.right = helper(nums, (start + end) / 2 + 1, end);
        return node;
    }
}

109. Convert Sorted List to Binary Search Tree

My Solutions:

和108类似，找到list的中点作为root。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode sortedListToBST(ListNode head) {
        if (head == null) return null;
        return helper(head, null);
    }
    
    public TreeNode helper(ListNode head, ListNode tail) {
        if (head == tail) return null;
        
        // 找到list中点
        ListNode slow = head, fast = head;
        while (fast != tail && fast.next != tail) {
            fast = fast.next.next;
            slow = slow.next;
        }
        TreeNode mid = new TreeNode(slow.val);
        mid.left = helper(head, slow);
        mid.right = helper(slow.next, tail);
        return mid;
    }
}