* 解题思路
* 1.先设置一个标志来判断是否是子树结构,如果是子结构停止遍历,如果不是继续遍历
* 2.调用trave(),如果根节点不同,判断左子树,相同判断右子树
* 当两棵树都不为空的时候才进行判断
* 当tree2为空,说明遍历tree2完成,tree2是tree1的子树
* 当tree1为空有如下情况
* (1)tree1和tree2都为空那么tree2是tree1的子树
* (2)tree1为空tree2不为空,那么tree2不是tree1的子树
public class Offer18 {
public static boolean HasSubtree(TreeNode root1,TreeNode root2){
boolean flag=false;
//判断根节点情况
if(root1.val==root2.val){
flag=Trave(root1,root2);
}
//根节点不相同去tree1左子树找
else if (!flag){
flag=HasSubtree(root1.left,root2);
}
//去右边找
else if (!flag){
flag=HasSubtree(root1.right,root2);
}
return flag;
}
public static boolean Trave(TreeNode root1,TreeNode root2){
if (root2==null){
return true;
}
else if(root1==null){
return false;
}
else if(root1.val!=root2.val){
return false;
}
else {
return Trave(root1.left,root2.left)&&Trave(root1.right,root2.right);
}
}
public static void main(String[] args) {
TreeNode wx=new TreeNode(1);
TreeNode wx1=new TreeNode(2);
TreeNode wx2=new TreeNode(3);
wx.left=wx1;
wx.right=wx2;
TreeNode w=new TreeNode(1);
TreeNode w1=new TreeNode(2);
TreeNode w2=new TreeNode(3);
w.left=w1;
w.right=w2;
boolean b = HasSubtree(wx, w);
System.out.println(b);
}
}