标签：
问题描述
将一棵二叉查找树(BST)转为有序的双向链表。
例如，有一颗BST如下：
2
| \
1 3
转成双向链表为：
1=2=3
解决思路
1. 保持有序：中序遍历；
2. 双向链表：记录链表的头节点，遍历过程中记录前一个节点并且保持双向连接关系。
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);
}
}
附加测试程序：
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;
}
}
}
时间/空间复杂度
时间复杂度：中序遍历的时间复杂度，O(n)；
空间复杂度：O(h)，递归栈。
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原文地址：http://www.cnblogs.com/harrygogo/p/4605841.html
