LeetCode 刷题【109. 有序链表转换二叉搜索树】

109. 有序链表转换二叉搜索树

自己做

解：左旋

/*** 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 {private int countTreeHeight(TreeNode root){if(root == null)return 0;int left = countTreeHeight(root.left);int right = countTreeHeight(root.right);return Integer.max(left, right) + 1;}public TreeNode sortedListToBST(ListNode head) {TreeNode root = null;TreeNode p = null;//插入根结点if(head != null){root = new TreeNode(head.val);p = root;head = head.next;}//后面边插入边调整while(head != null){p.right = new TreeNode(head.val);p = p.right;//插入后破坏平衡，检查有无破坏平衡//找到破坏平衡的最小子树【对应树的根结点为last_q，对应子树的父母为last】TreeNode last = null;TreeNode last_q = null;TreeNode q = root;   while(q != null){int left = countTreeHeight(q.left);int right = countTreeHeight(q.right);  //破坏了平衡的情况if(right - left >= 2){//找最小破坏平衡的子树while(right - left >= 2){last = last_q;last_q = q;q = q.right;left = countTreeHeight(q.left);right = countTreeHeight(q.right);  }//左旋调整if(last_q == root){            //要调整根结点的情况last_q.right = q.left;q.left = last_q;root = q;}else{                          //不用调整根结点的情况last_q.right = q.left;q.left = last_q;                    last.right = q;}//重置检查last = null;last_q = null;q = root;  }//没有破坏平衡，继续往下找else{last = last_q;last_q = q;q = q.right;}}head = head.next;}return  root;}
}

看题解

解：分治

官方题解：

class Solution {ListNode globalHead;public TreeNode sortedListToBST(ListNode head) {globalHead = head;int length = getLength(head);return buildTree(0, length - 1);}public int getLength(ListNode head) {int ret = 0;while (head != null) {++ret;head = head.next;}return ret;}public TreeNode buildTree(int left, int right) {if (left > right) {return null;}int mid = (left + right + 1) / 2;TreeNode root = new TreeNode();root.left = buildTree(left, mid - 1);root.val = globalHead.val;globalHead = globalHead.next;root.right = buildTree(mid + 1, right);return root;}
}

啊哈哈哈哈哈，原来根本不用左旋右旋这些