《剑指Offer》36：二叉搜索树与双向链表

题目

输入一棵二叉搜索树，将该二叉搜索树转换成一个排序的双向链表。要求不能创建任何新的节点，只能调整树中节点指针的指向。比如，输入下图中的二叉搜索树，输出转换之后的排序双向链表。

二叉树节点的定义如下：

public static class TreeNode {
	public int val;
	public TreeNode left;
	public TreeNode right;
	public TreeNode(int x) { val = x; }
}

分析

众所周知，中序遍历二叉搜索树会得到有序的序列，我们目标是在中序遍历二叉搜索树过程中，逐步将其转换成有序的双向链表。另外，将树节点的左子树指针转换成双向链表节点的前驱指针，而树节点的右子树指针转换成双向链表节点的后驱指针。

放码

import com.lun.util.BinaryTree.TreeNode;

public class ConvertBSTToLinkedList {

	private TreeNode last;//用于指向双向链表的尾节点
	
	public TreeNode convert(TreeNode root) {
		convertNode(root);
		
		TreeNode head = last;
		while(head != null && head.left != null) {
			head = head.left;
		}
		return head;
	}

	private void convertNode(TreeNode node) {
		
		if(node == null) {
			return;
		}
		
		TreeNode current = node;
		
		if(current.left != null) {
			convertNode(current.left);
		}
		
		current.left = last;//1.执行到这步，左子树已经转换成有序双向链表
		
		if(last != null) {
			last.right = current;//2.
		}
		
		last = current;//3.current转换成有序双向链表的新尾节点
		
		if(current.right != null) {
			convertNode(current.right);
		}
	}
	
}

测试

import org.junit.Assert;
import org.junit.Test;

import com.lun.util.BinaryTree;
import com.lun.util.BinaryTree.TreeNode;

public class ConvertBSTToLinkedListTest {

	@Test
	public void test() {
		ConvertBSTToLinkedList cbl = new ConvertBSTToLinkedList();
		TreeNode root = makeABST();
		TreeNode head = cbl.convert(root);
		Assert.assertEquals("4 -> 6 -> 8 -> 10 -> 12 -> 14 -> 16 -> \n" + 
				"16 -> 14 -> 12 -> 10 -> 8 -> 6 -> 4 -> ", printList(head));
	}
	
	private TreeNode makeABST() {
		int[] array = {10, 6, 14, 4, 8, 12, 16};
		return BinaryTree.integerArray2BinarySearchTree(array);
	}
	
	private String printList(TreeNode head) {
		String result = "";
		
		TreeNode p = head;
		
		while(true) {
			result += (p.val + " -> ");
			
			if(p.right == null) {
				break;
			}
			p = p.right;
		}

		result += "\n";
		
		while(p != null) {
			result = result +  p.val + " -> ";
			p = p.left;
		}
		return result;
	}
	
}