[LeetCode#109]Convert Sorted List to Binary Search Tree

The problem: 

Convert Sorted List to Binary Search Tree

 

Link:

https://oj.leetcode.com/problems/convert-sorted-list-to-binary-search-tree/

 

My analysis:

The idea of building a binary tree from a sorted linked list is same as from a sorted array.
The key issues here is:
1. locate the mid element of linked list
2. split the linked list into two parts according mid element.

The classic way of finding the mid element on a linked list is to use two pointers: walker and runner.

while (runner.next != null && runner.next.next != null) {
            
            pre_walker = walker;
            walker = walker.next;
            runner = runner.next.next;
}

Note: the mid element would be the node pointer by walker, to split the list into two parts, we have to keep a pre_walker pointer to point the previous element of walker.

One pitfall.
The corner case: when there are only two elements in the list.
In this situation, we take the first element as root, the second element as the right child.
In this case, the above invariant could not be used, we should tackle it specically. And, we could detect this case by testing if pre_walker has already been initialized.

if (pre_walker == null) {
    head1 = null;
} else { 
    head1 = head;
    pre_walker.next = null; 
}

Then the problem was fixed!

 

My solution:

public class Solution {
    public TreeNode sortedListToBST(ListNode head) {
        
        if (head == null)
            return null;
            
        return helper(head);
    }
    
    private TreeNode helper(ListNode head) {
    
        if (head == null)
            return null;

        if (head.next == null)
            return new TreeNode(head.val);
            
        ListNode head1;
        ListNode head2;
        ListNode walker = head;
        ListNode runner = head;
        ListNode pre_walker = null; //use to record the pre node of walker pointer
        ListNode mid;

        while (runner.next != null && runner.next.next != null) {
            
            pre_walker = walker;
            walker = walker.next;
            runner = runner.next.next;
        }
        
        mid = walker; //the mid position is pointed by walker
        TreeNode ret = new TreeNode(mid.val);
        
        /*
        if pre_walker has already been initialized, there are at list three elements left in the list.
        */
        if (pre_walker == null) { //careful! only two elements left, we choose the head as mid element.
            head1 = null;
        } else { 
            head1 = head;
            pre_walker.next = null; //split the original linked list at walker. 
        }
        
        head2 = mid.next;
        ret.left = helper(head1);
        ret.right = helper(head2);
        
        return ret; 
    }
}

 

转载于:https://www.cnblogs.com/airwindow/p/4209797.html