bstToDoublyList

bstToDoublyList

Table of Contents

• 1. 描述
• 2. 样例
• 3. 解决方案

1 描述

将一个二叉查找树按照中序遍历转换成双向链表。

2 样例

给定一个二叉查找树：

    4
   / 
  2   5
 / 
1   3

返回 1<->2<->3<->4<->5 。

3 解决方案

做一个中序递归，用head保存首节点，用tail不断更新之后的每个节点。

/**
 * 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 {
    private DoublyListNode head = null;
    private DoublyListNode tail = null;
    /**
     * @param root: The root of tree
     * @return: the head of doubly list node
     */
    public DoublyListNode bstToDoublyList(TreeNode root) {
        // Write your code here
        if (null == root) {
            return null;
        }
        bstToDoublyList(root.left);

        DoublyListNode node = new DoublyListNode(root.val);
        if (null == head) {
            head = node;
            tail = node;
        } else {
            tail.next = node;
            node.prev = tail;
            tail = node;
        }

        bstToDoublyList(root.right);

        return head;
    }
}

Date: 2017-06-26 21:57

Author: WEN YANG

Created: 2017-06-26 Mon 22:03

Emacs 25.2.1 (Org mode 8.2.10)

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