[LeetCode]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.

Example

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Recursion Linked List

解题思路：

思路一：平衡树调整。第一步建立一个单向树，然后使用平衡树的调整。逻辑比较绕，虽然完成了代码，但是大数据时，测试超时。

思路二：保存所有节点到数组（List集合），然后使用二分法，依次建立所有的头节点和左子树，右子树。

思路三：参考网上解法，核心还是使用二分法，但是比较巧妙的一点是在确定所有元素个数和划分左右子树后，能够实现指针的一次遍历，读取全部数据，无需多次将数据读出保存。

思路一：

/** * Definition for ListNode. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int val) { *         this.val = val; *         this.next = null; *     } * } * Definition of TreeNode: * public class TreeNode { *     public int val; *     public TreeNode left, right; *     public TreeNode(int val) { *         this.val = val; *         this.left = this.right = null; *     } * } */ public class Solution {    /**     * @param head: The first node of linked list.     * @return: a tree node     */    public TreeNode sortedListToBST(ListNode head) {if (head == null)return null;TreeNode root = null;while (head != null) {int v = head.val;head = head.next;TreeNode node = new TreeNode(v);node.left = root;root = node;}return adjustNode(root);}TreeNode adjustNode(TreeNode node) {if (node != null) {int l = getLevel(node.left);int r = getLevel(node.right);while (l - r >= 2) {TreeNode tmp = node.left;if (tmp.right == null) {node.left = null;tmp.right = node;node = tmp;} else {TreeNode lmax = tmp.right;while (lmax.right != null) {tmp = lmax;lmax = lmax.right;}tmp.right = null;lmax.left = node.left;lmax.right = node;node.left = null;node = lmax;}l = getLevel(node.left);r = getLevel(node.right);}while (r - l >= 2) {TreeNode tmp = node.right;if (tmp.left == null) {node.right = null;tmp.left = node;node = tmp;} else {TreeNode lmax = tmp.left;while (lmax.left != null) {tmp = lmax;lmax = lmax.left;}tmp.left = null;lmax.right = node.right;lmax.left = node;node.right = null;node = lmax;}l = getLevel(node.left);r = getLevel(node.right);}node.left = adjustNode(node.left);node.right = adjustNode(node.right);}return node;}int getLevel(TreeNode node) {if (node == null)return 0;elsereturn Math.max(getLevel(node.left) + 1, getLevel(node.right) + 1);}}

解法二：

/** * Definition for ListNode. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int val) { *         this.val = val; *         this.next = null; *     } * } * Definition of TreeNode: * public class TreeNode { *     public int val; *     public TreeNode left, right; *     public TreeNode(int val) { *         this.val = val; *         this.left = this.right = null; *     } * } */ public class Solution {    /**     * @param head: The first node of linked list.     * @return: a tree node     */    public TreeNode sortedListToBST(ListNode head) {if (head == null)return null;ArrayList<Integer> list = new ArrayList<Integer>();while (head != null) {list.add(head.val);head = head.next;}TreeNode root = getNode(list, 0, list.size() - 1);return root;}private TreeNode getNode(ArrayList<Integer> list, int x1, int x2) {if (x1 == x2)return new TreeNode(list.get(x1));else if (x1 > x2)return null;else {int mid = (x1 + x2) / 2;TreeNode node = new TreeNode(list.get(mid));node.left = getNode(list, x1, mid - 1);node.right = getNode(list, mid + 1, x2);return node;}}}

思路三：（网上解法）

public TreeNode sortedListToBST(ListNode head) {      if(head == null)          return null;      ListNode cur = head;      int count = 0;      while(cur!=null)      {          cur = cur.next;          count++;      }      ArrayList<ListNode> list = new ArrayList<ListNode>();      list.add(head);      return helper(list,0,count-1);  }  private TreeNode helper(ArrayList<ListNode> list, int l, int r)  {      if(l>r)          return null;      int m = (l+r)/2;      TreeNode left = helper(list,l,m-1);      TreeNode root = new TreeNode(list.get(0).val);      root.left = left;      list.set(0,list.get(0).next);      root.right = helper(list,m+1,r);      return root;  }