标签：param class convert logs return blog bin roo 实现
Convert a binary search tree to doubly linked list with in-order traversal.
Example
Given a binary search tree:
4
/ 2 5
/ 1 3
return 1<->2<->3<->4<->5
对每一个根节点来说，就是左边连上左子树的last(右下角)，右边连上右子树的first(左下角)。然后返回自己的first(左下角)和last(右下角)。
用resultType来返回first last。
注意单侧null的细节处理。
1.自己实现
/** * 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; * } * } * Definition for Doubly-ListNode. * public class DoublyListNode { * int val; * DoublyListNode next, prev; * DoublyListNode(int val) { * this.val = val; * this.next = this.prev = null; * } * } */ public class Solution { /* * @param root: The root of tree * @return: the head of doubly list node */ private class ResultType{ public DoublyListNode first; public DoublyListNode last; public ResultType(DoublyListNode first, DoublyListNode last) { this.first = first; this.last = last; } } public DoublyListNode bstToDoublyList(TreeNode root) { // write your code here if (root == null) { return null; } ResultType result = helper(root); return result.first; } private ResultType helper(TreeNode root) { if (root == null) { return null; } DoublyListNode rootNode = new DoublyListNode(root.val); if (root.left == null && root.right == null) { return new ResultType(rootNode, rootNode); } ResultType left = helper(root.left); ResultType right = helper(root.right); if (left != null) { left.last.next = rootNode; rootNode.prev = left.last; } if (right != null) { rootNode.next = right.first; right.first.prev = rootNode; } DoublyListNode leftFirst = left == null ? rootNode : left.first; DoublyListNode rightLast = right == null ? rootNode : right.last; return new ResultType(leftFirst, rightLast); } }
2. 九章实现：把 left == null 和left != null写到一个队里if clause里，更清楚，学习
但大体上这次我是自己写的答案和题解最接近的一次了！！！真的非常开心，而且一次AC。
class ResultType { DoublyListNode first, last; public ResultType(DoublyListNode first, DoublyListNode last) { this.first = first; this.last = last; } } public class Solution { /** * @param root: The root of tree * @return: the head of doubly list node */ public DoublyListNode bstToDoublyList(TreeNode root) { if (root == null) { return null; } ResultType result = helper(root); return result.first; } public ResultType helper(TreeNode root) { if (root == null) { return null; } ResultType left = helper(root.left); ResultType right = helper(root.right); DoublyListNode node = new DoublyListNode(root.val); ResultType result = new ResultType(null, null); if (left == null) { result.first = node; } else { result.first = left.first; left.last.next = node; node.prev = left.last; } if (right == null) { result.last = node; } else { result.last = right.last; right.first.prev = node; node.next = right.first; } return result; } }
lintcode378- Convert Binary Search Tree to Doubly Linked Lis- medium
标签：param class convert logs return blog bin roo 实现
原文地址：http://www.cnblogs.com/jasminemzy/p/7684861.html
