18 树的子结构

package question18;

import java.util.Scanner;

class TreeNode{
int val;
TreeNode left=null;
TreeNode right=null;
public TreeNode(int val) {
this.val=val;
}
}
public class SubstructureInTree {

public static void main(String[] args) {

        // TODO Auto-generated method stub

        Scanner in=new Scanner(System.in);

        System.out.println("请输入树 1 的先序遍历:");
        TreeNode root1=creatTree(in);
        System.out.println("下面是树 1 的先序遍历:");
        display(root1);
        System.out.println();
        System.out.println("请输入树 2 的先序遍历:");
        TreeNode root2=creatTree(in);
        System.out.println("下面是树 2 的先序遍历:");
        display(root2);
        System.out.println();

        boolean result=    HasSubtree(root1,root2);
        System.out.println("树 1 是否包含 树 2:  "+result);
   

}

//树1 是否包含 树2
private static boolean HasSubtree(TreeNode root1, TreeNode root2) {
// TODO Auto-generated method stub
boolean result=false;
if (root1!=null&&root2!=null) {
if (root1.val==root2.val) {
result=DoesTreeHaveTree2(root1,root2);
}
if (!result) {
result=HasSubtree(root1.left, root2);
}
if (!result) {
result=HasSubtree(root1.right, root2);
}
}
return result;

}
// 判断 ：当根节点相同时，树1 是否包含 树2
private static boolean DoesTreeHaveTree2(TreeNode root1, TreeNode root2) {
// TODO Auto-generated method stub
if (root2==null ) {
return true;
}

if (root1==null) {
return false;
}

if (root1.val!=root2.val) {
return false;
}

return DoesTreeHaveTree2(root1.left, root2.left)&&DoesTreeHaveTree2(root1.right, root2.right);
}
// 打印二叉树
private static void display(TreeNode root) {
// TODO Auto-generated method stub

if (root!=null) {
System.out.print(root.val+"  ");
display( root.left);
display(root.right);

}
}
// 创建二叉树
private static TreeNode creatTree(Scanner in) {
// TODO Auto-generated method stub
TreeNode newNode;
int val=in.nextInt();
if (val==-1) {     //设置 -1 为结束数字
val=in.nextInt();
}
if (val!=0) {    // 0 代表空节点
newNode=new TreeNode(val);
}else {
newNode=null;
return null;
}

newNode.left=creatTree(in);
newNode.right=creatTree(in);

return newNode;