問題描述

將一棵二叉查找樹(BST)轉(zhuǎn)為有序的雙向鏈表。

例如，有一顆BST如下：

2

| ?\

1 ? ?3

轉(zhuǎn)成雙向鏈表為：

1=2=3

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解決思路

1. 保持有序：中序遍歷；

2. 雙向鏈表：記錄鏈表的頭節(jié)點(diǎn)，遍歷過程中記錄前一個節(jié)點(diǎn)并且保持雙向連接關(guān)系。

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程序

class TreeNode {public int val;public TreeNode left;public TreeNode right;public TreeNode(int _val) {val = _val;} }class BSTToDoublyLinkedList {public TreeNode convertToDoublyLinkedList(TreeNode root) {if (root == null) {return null;}TreeNode head = new TreeNode(0);TreeNode pre = new TreeNode(0);convertHelper(root, head, pre);return head.left;}private void convertHelper(TreeNode root, TreeNode head, TreeNode pre) {if (root == null) {return;}convertHelper(root.left, head, pre);if (head.left == null) {head.left = root; // record the head} if (pre.left == null) {pre.left = root;} else {pre.left.right = root;root.left = pre.left;pre.left = root; // to next}convertHelper(root.right, head, pre);} }

　　

附加測試程序：

public class ConvertTest {public static void main(String[] args) {TreeNode n2 = new TreeNode(3);n2.left = new TreeNode(1);n2.left.right = new TreeNode(2);BSTToDoublyLinkedList toDoublyLinkedList = new BSTToDoublyLinkedList();TreeNode head = toDoublyLinkedList.convertToDoublyLinkedList(n2);printDoublyLinkedList(head);}private static void printDoublyLinkedList(TreeNode head) {TreeNode node = head;TreeNode last = null;System.out.println("left --> right");while (node != null) {System.out.print(node.val + " ");if (node.right == null) {last = node;}node = node.right;}System.out.println();System.out.println("right --> left");while (last != null) {System.out.print(last.val + " ");last = last.left;}} }

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時間/空間復(fù)雜度

時間復(fù)雜度：中序遍歷的時間復(fù)雜度，O(n)；

空間復(fù)雜度：O(h)，遞歸棧。

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轉(zhuǎn)載于:https://www.cnblogs.com/harrygogo/p/4605841.html

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