第15 题：
题目：输入一颗二元查找树，将该树转换为它的镜像，
即在转换后的二元查找树中，左子树的结点都大于右子树的结点。
用递归和循环两种方法完成树的镜像转换。
例如输入：
8
/
6 10
/ /
5 7 9 11
输出：
8
/
10 6
/ /
11 9 7 5
思路：
1.递归： 根节点的镜像必然由其左子树的镜像和右子树的镜像组成，并将左子树的镜像置于根节点的右边，右子树的镜像放于根节点的左边，拿到左右子树，分别使用上述逻辑对其递归即可，递归的结束条件为如节点为空则返回null，如果是子节点则返回该子节点
2.非递归： 从根节点开始，依次swap其左右分支节点，然后依次处理左右分支节点，因为要保留从根节点到左右分支节点的路径，可以用队列来完成这个逻辑, 队列的头结点为本次要处理的节点，将其出队列，并将其左右节点分别入队列，以便稍后处理
1 package com.rui.microsoft; 2 3 import java.util.ArrayDeque; 4 import java.util.Deque; 5 6 public class Test15_BSTMirror { 7 8 public static void main(String[] args) { 9 Node node4 = new Node(4); 10 Node node3 = new Node(3); 11 Node node5 = new Node(5); 12 Node node2 = new Node(2); 13 14 node4.left = node3; 15 node4.right = node5; 16 node3.left = node2; 17 18 midOrder(node4); 19 Node head = mirrorRec(node4); 20 //Node head = mirrorNonRec(node4); 21 System.out.println(" "); 22 midOrder(head); 23 } 24 25 //Recursive 26 public static Node mirrorRec(Node node){ 27 Node head = node; 28 if(null == head) return null; 29 //leaf 30 if(null == head.left && null == head.right) return head; 31 32 Node left = head.left; 33 Node right = head.right; 34 head.right = mirrorRec(left); 35 head.left = mirrorRec(right); 36 return head; 37 } 38 39 //Non-recursive 40 public static Node mirrorNonRec(Node node){ 41 Node head = node; 42 43 if(null == head) return null; 44 //leaf 45 if(null == head.left && null == head.right) return head; 46 47 //Non-leaf 48 Deque<Node> queue = new ArrayDeque<Node>(); 49 queue.offer(head); 50 51 while(!queue.isEmpty()){ 52 Node n = queue.poll(); 53 Node left = n.left; 54 Node right = n.right; 55 if(null != left)queue.offer(left); 56 if(null != right)queue.offer(right); 57 n.left = right; 58 n.right = left; 59 } 60 61 return head; 62 } 63 64 public static void midOrder(Node head){ 65 if(null == head) return; 66 midOrder(head.left); 67 System.out.print(" " + head.value); 68 midOrder(head.right); 69 } 70 71 static class Node{ 72 int value; 73 Node left; 74 Node right; 75 Node(int v){ 76 this.value = v; 77 } 78 } 79 }