LeetCode - Convert Sorted List to Binary Search Tree
Problem description
Given a singly linked list 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:
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Given the sorted linked list: [-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
Analysis
General idea is to use inorder traversal. But we need a input to control the traversal, either the TreeNode itself or the left and right boundary.
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/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
/**
* 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 {
ListNode head;
public TreeNode sortedListToBST(ListNode head) {
ListNode p = head;
this.head = head;
int size = 0;
while(p!=null){
p =p.next;
size++;
}
return build(0, size - 1);
}
TreeNode build(int l, int h){
if (l > h){
return null;
}
int mid = l + (h-l) / 2;
TreeNode left = build(l, mid - 1);
TreeNode node = new TreeNode(head.val);
head = head.next;
TreeNode right = build(mid+1, h);
node.left = left;
node.right = right;
return node;
}
}
What to improve
- when using traversal to build a tree, we can use low and high as the input of the traversal.