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https://oj.leetcode.com/problems/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.
解题思路：
首先，这题与 Convert Sorted Array to Binary Search Tree 很类似，可以用递归的方法来解。难度在于，List和Array的不同，不能直接获取节点的下标，以及计算这段list的中点，也不能比较他们start和end下标的大小，以便在start>end的时候合理退出。所以需要取得midNode，同时还要获得midNode的前一个节点。每次递归进入方法的时候，需要对于只有一个节点，和只有两个节点的情况，分别处理，否则就会出错。
/** * 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) { if(head == null){ return null; } ListNode end = head; while(end.next != null){ end = end.next; } return sortedListToBSTHelper(head, end); } public TreeNode sortedListToBSTHelper(ListNode head, ListNode end){ //nodeNum == 0，只有一个节点的情况 if(head == end){ return new TreeNode(head.val); } int nodeNum = 0; ListNode tempNode = head; while(tempNode != end){ tempNode = tempNode.next; nodeNum++; } //nodeNum == 1，只有两个节点的情况 if(nodeNum == 1){ TreeNode root = new TreeNode(head.val); root.right = sortedListToBSTHelper(head.next, end); return root; } //nodeNum > 2的情况 TreeNode preMidNode = head; for(int i = 0; i < nodeNum / 2 - 1; i++){ preMidNode = preMidNode.next; } ListNode midNode = preMidNode.next; TreeNode root = new TreeNode(midNode.val); root.left = sortedListToBSTHelper(head, preMidNode); root.right = sortedListToBSTHelper(midNode.next, end); return root; } }
后来在网上看到一个左闭右开的方法，类似于二分查找，在 Find Minimum in Rotated Sorted Array 和 Search Insert Position 问题里提到过这个问题。那么右边界只要取midNode就可以了，不必取得他的钱一个节点。同时，调用递归方法的时候，也必须为sortedListToBSTHelper(head, null);而不能是sortedListToBSTHelper(head, end)。
还要注意，计算列表节点数量的方法也发生了变化，因为end已经是一个开放的区间。
这样避免了preMidNode节点和start的判断问题，代码简单了。
/** * 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) { if(head == null){ return null; } return sortedListToBSTHelper(head, null); } public TreeNode sortedListToBSTHelper(ListNode head, ListNode end){ if(head == end){ return null; } //nodeNum == 0，只有一个节点的情况 if(head.next == end){ return new TreeNode(head.val); } int nodeNum = 0; ListNode tempNode = head; while(tempNode.next != end){ tempNode = tempNode.next; nodeNum++; } //nodeNum == 1，只有两个节点的情况，可以省略 //左闭右开的方法下面已经涵盖 // if(nodeNum == 1){ // TreeNode root = new TreeNode(head.val); // root.right = sortedListToBSTHelper(head.next, end); // return root; // } //nodeNum > 2的情况 ListNode midNode = head; for(int i = 0; i < nodeNum / 2; i++){ midNode = midNode.next; } TreeNode root = new TreeNode(midNode.val); root.left = sortedListToBSTHelper(head, midNode); root.right = sortedListToBSTHelper(midNode.next, end); return root; } }
把取得中间节点作为一个方法拿出来，refactor代码。
/** * 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) { if(head == null){ return null; } return sortedListToBSTHelper(head, null); } public TreeNode sortedListToBSTHelper(ListNode head, ListNode end){ if(head == end){ return null; } //nodeNum == 0，只有一个节点的情况 if(head.next == end){ return new TreeNode(head.val); } ListNode midNode = getMidNode(head, end); TreeNode root = new TreeNode(midNode.val); root.left = sortedListToBSTHelper(head, midNode); root.right = sortedListToBSTHelper(midNode.next, end); return root; } public ListNode getMidNode(ListNode head, ListNode end){ int nodeNum = 0; ListNode tempNode = head; while(tempNode.next != end){ tempNode = tempNode.next; nodeNum++; } ListNode midNode = head; for(int i = 0; i < nodeNum / 2; i++){ midNode = midNode.next; } return midNode; } }
利用快慢指针的方法，找一个链表中点，可以写成下面的方法。
public ListNode getMidNode(ListNode head, ListNode end){ ListNode quickNode = head; ListNode slowNode = head; while(quickNode != end && quickNode.next != end){ quickNode = quickNode.next.next; slowNode = slowNode.next; } return slowNode; } }
或者下面的形式，更容易理解一些。
public ListNode getMidNode(ListNode head, ListNode end){ ListNode quickNode = head; ListNode slowNode = head; while(quickNode.next != end){ quickNode = quickNode.next; if(quickNode.next != end){ quickNode = quickNode.next; slowNode = slowNode.next; } } return slowNode; } }
同样下面的方法也是对的，想想为什么循环终止的条件，quickNode != end 和 quickNode.next != end都可以？
public ListNode getMidNode(ListNode head, ListNode end){ ListNode quickNode = head; ListNode slowNode = head; while(quickNode != end){ quickNode = quickNode.next; if(quickNode != end){ quickNode = quickNode.next; slowNode = slowNode.next; } } return slowNode; }
因为1-2-3-4这样的数列，下面两种BST都是平衡的。
3 2
/ \ / \
2 4 1 3
/ \
1 4
Convert Sorted List to Binary Search Tree
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原文地址：http://www.cnblogs.com/NickyYe/p/4310355.html
