109. Convert Sorted List to Binary Search Tree

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.

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 需要总长度，遍历一遍，

1. 可以再这个时候吧list转成array， 费地方

2. 或者跟array一样每次走半个长度找到mid， O(NlogN), logN层，每层遍历N，费时间

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; next = null; }
 * }
 */
/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode sortedListToBST(ListNode head) {
        
        // traverse the get the length
        ListNode node = head;
        int len = 0;
        while(node != null){
            node = node.next;
            len++;
        }
        
        return helper(head, 0, len-1);
        
    }
    
    private TreeNode helper(ListNode node, int l, int r){
        if(l>r)     return null;
        
        int mid = l + (r-l)/2;
        
        // find the mid
        int i=0;
        ListNode run = node;
        while(i<mid){
            run = run.next;
            i++;
        }
        
        TreeNode cur = new TreeNode(run.val);
        
        cur.left = helper(node, l, mid-1);  
        cur.right = helper(node, mid+1, r);
        
        return cur;        
    }
}

 

3. bottem up, 先从left child-》cur -> right, 按list node 插入，

但是注意object passed by reference, 本层node = node.next,但是上一层的node还是没变

所以用node.val = node.next val, node.next = node.next.next直接改变reference所指的node的值

O(N)

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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; next = null; }
 * }
 */
/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode sortedListToBST(ListNode head) {
        
        // traverse the get the length
        ListNode node = head;
        int len = 0;
        while(node != null){
            node = node.next;
            len++;
        }
        
        return helper(head, 0, len-1);
        
    }
    
    private TreeNode helper(ListNode node, int l, int r){
        if(l>r)     return null;
        
        int mid = l + (r-l)/2;
        // left first
        TreeNode left = helper(node, l, mid-1);
        TreeNode cur = new TreeNode(node.val);
        cur.left = left;
        
        // java pass object by reference,
        // even if i call node = node.next;
        // the node the other recursive call dosn't change
        // so move the node by :
        
        if(node.next != null){
            node.val = node.next.val;
            node.next = node.next.next;            
        }
        
        
        cur.right = helper(node, mid+1, r);
        return cur;        
    }
}