/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode reConstructBinaryTree(int [] pre,int [] in) {
        TreeNode root=reConstructBinaryTree(pre,0,pre.length-1,in,0,in.length-1);
        return root;
    }
    
    public TreeNode reConstructBinaryTree(int[] pre,int preStart,int preEnd,int[] in,int inStart,int inEnd){
        if(preStart>preEnd||inStart>inEnd){
            return null;
        }
        
        TreeNode root=new TreeNode(pre[preStart]);
        for(int i=inStart;i<=inEnd;i++){
            if(pre[preStart]==in[i]){
                root.left=reConstructBinaryTree(pre,preStart+1,preStart+i-inStart,in,inStart,i-1);
                root.right=reConstructBinaryTree(pre,preStart+i-inStart+1,preEnd,in,i+1,inEnd);
                break;
            }
        }
        return root;
    }
}

public class Solution {
    public static boolean HasSubtree(TreeNode root1, TreeNode root2) {
        boolean result = false;
        //当Tree1和Tree2都不为零的时候，才进行比较。否则直接返回false
        if (root2 != null && root1 != null) {
            //如果找到了对应Tree2的根节点的点
            if(root1.val == root2.val){
                //以这个根节点为为起点判断是否包含Tree2
                result = doesTree1HaveTree2(root1,root2);
            }
            //如果找不到，那么就再去root的左儿子当作起点，去判断时候包含Tree2
            if (!result) {
                result = HasSubtree(root1.left,root2);
            }
            
            //如果还找不到，那么就再去root的右儿子当作起点，去判断时候包含Tree2
            if (!result) {
                result = HasSubtree(root1.right,root2);
               }
            }
            //返回结果
        return result;
    }

    public static boolean doesTree1HaveTree2(TreeNode node1, TreeNode node2) {
        //如果Tree2已经遍历完了都能对应的上，返回true
        if (node2 == null) {
            return true;
        }
        //如果Tree2还没有遍历完，Tree1却遍历完了。返回false
        if (node1 == null) {
            return false;
        }
        //如果其中有一个点没有对应上，返回false
        if (node1.val != node2.val) {   
                return false;
        }
        
        //如果根节点对应的上，那么就分别去子节点里面匹配
        return doesTree1HaveTree2(node1.left,node2.left) && doesTree1HaveTree2(node1.right,node2.right);
    }
}

import java.util.*;

public class Solution {
    public TreeNode Mirror (TreeNode pRoot) {
        if(pRoot==null)
            return pRoot;
        if(pRoot.left==null && pRoot.right==null)
            return pRoot;
        //处理根节点，交换左右节点
        TreeNode temp=pRoot.left;
        pRoot.left=pRoot.right;
        pRoot.right=temp;
        //处理左子树
        Mirror(pRoot.left);
        //处理右子树
        Mirror(pRoot.right);
        return pRoot;
    }
}