剑指offer面试题27：二叉搜索树与双向链表

题目：输入一颗二叉搜索树，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。

由于二叉搜索树是有序的，左子结点的值小于根节点的值，右子结点的值大于根节点的值。所以在把二叉搜索树转换成排序的双向链表的时候要把左子树中的最大值的右子树指针指向根节点，把右子树中的最小值的左子树指针指向根节点。

由于先访问根节点，因此要用中序遍历的方式进行处理。

package Solution;

public class No27ConvertBinarySearchTreeToLinkedList {

static class BinaryTreeNode {
int value;
BinaryTreeNode left;
BinaryTreeNode right;

public BinaryTreeNode() {

}

public BinaryTreeNode(int value, BinaryTreeNode left,
BinaryTreeNode right) {
this.value = value;
this.left = left;
this.right = right;
}
}

public static BinaryTreeNode convert(BinaryTreeNode root) {
BinaryTreeNode lastNodeInList = null;
lastNodeInList = convertToNode(root, lastNodeInList);
BinaryTreeNode head = lastNodeInList;
// 从尾节点返回头结点
while (head != null && head.left != null) {
head = head.left;
}
printList(head);
return head;
}

private static BinaryTreeNode convertToNode(BinaryTreeNode node,
BinaryTreeNode lastNodeInList) {
if (node == null)
return null;
BinaryTreeNode current = node;
// 递归的处理左子树
if (current.left != null)
lastNodeInList = convertToNode(current.left, lastNodeInList);
// 使链表中的最后一个结点指向左子树的最小的节点
current.left = lastNodeInList;
// 链表中的最后一个结点指向当前节点，当前节点就成了链表中的最后一个结点
if (lastNodeInList != null)
lastNodeInList.right = current;
lastNodeInList = current;
// 递归转换右子树
if (current.right != null)
lastNodeInList = convertToNode(current.right, lastNodeInList);

return lastNodeInList;
}

public static void printList(BinaryTreeNode head) {
while (head != null) {
System.out.print(head.value + ",");
head = head.right;
}
}

// 中序遍历二叉树
public static void printTree(BinaryTreeNode root) {
if (root != null) {
printTree(root.left);
System.out.print(root.value + ",");
printTree(root.right);
}
}

public static void main(String[] args) {
BinaryTreeNode node1 = new BinaryTreeNode();
BinaryTreeNode node2 = new BinaryTreeNode();
BinaryTreeNode node3 = new BinaryTreeNode();
BinaryTreeNode node4 = new BinaryTreeNode();
BinaryTreeNode node5 = new BinaryTreeNode();
BinaryTreeNode node6 = new BinaryTreeNode();
BinaryTreeNode node7 = new BinaryTreeNode();
node7.value = 16;
node6.value = 12;
node5.value = 14;
node5.left = node6;
node5.right = node7;
node3.value = 4;
node4.value = 8;
node2.value = 6;
node2.left = node3;
node2.right = node4;
node1.value = 10;
node1.left = node2;
node1.right = node5;
printTree(node1);
System.out.println();
System.out.println("=============打印链表================");
convert(node1);
}
}