二叉树-Convert Sorted List to Binary Search Tree

2014.8.18 update:

其实相当简单，就是找到链表的中间node然后左半边成为left node, 右半边成为right node

public class Solution {    public TreeNode sortedListToBST(ListNode head) {        if (head == null) {            return null;        }        ListNode tail = head;        while (tail.next != null) {            tail = tail.next;        }                return helper(head, tail);    }        TreeNode helper(ListNode head, ListNode tail) {        if (head == null || tail == null) {            return null;        }        if (head == tail) {            return new TreeNode(head.val);        }                ListNode fast = head;        ListNode slow = head;        ListNode preSlow = null;        while (fast.next != null && fast != tail) {            fast = fast.next;            if (fast.next == null || fast == tail) {                break;            } else {                fast = fast.next;            }            preSlow = slow;            slow = slow.next;        }                TreeNode returnNode = new TreeNode(slow.val);        returnNode.left = helper(head, preSlow);        returnNode.right = helper(slow.next, tail);        return returnNode;    }}

2013:

public class Solution {

    public TreeNodesortedListToBST(ListNode head) {

       // Start typing your Java solution below

       // DO NOT write main() function

       if (head == null) {

           returnnull;

       }

       ListNode cur = head;

       int count = 0;

       while (cur != null) {

          count++;

           cur =cur.next;

       }

       return helper(head, 0, count-1);

    }

    TreeNode helper(ListNodelistHead, int low, int high) {

       if (low > high) {

           returnnull;

       }

       int mid = low + (high-low)/2;

       TreeNode left = helper(listHead, low,mid-1);

       TreeNode parent = newTreeNode(listHead.val);

       if (listHead.next != null) {

           listHead.val = listHead.next.val;

           listHead.next = listHead.next.next;

       }

       TreeNode right = helper(listHead, mid+1,high);

       parent.left = left;

       parent.right = right;

       return parent;

    }

}