方法一：最简单就是首先判断两棵树的根节点是否相同，如果相同则判断两颗树的左节点是否对应，右结点是否对应，

如果两个根节点值不同，主树左节点与子树根节点判断，主树右结点与子树根节点判断，然后再判断对应节 点是否相同即可

复杂度：O(m*n)

方法二：首先两棵树序列化为字符串，然后就是判断字符串之间的包含问题了，用KMP算法判断结果

复杂度：O(m+n)

上代码：

<span style="font-size:18px;">public class Solution {
    public boolean HasSubtree(TreeNode root1,TreeNode root2) {
        if(root1 == null&&root2 != null)return false;
        if(root2 == null)return false;
        boolean flag = false;
        if(root1 != null&&root2 != null){
            if(root1.val == root2.val){
            	flag = IsSubtree(root1,root2);
        	}
            if(!flag){
                flag = HasSubtree(root1.left,root2);
            }
            if(!flag){
                flag = HasSubtree(root1.right,root2);
            }
        }
        return flag;
    }
    public boolean IsSubtree(TreeNode root1,TreeNode root2){
        if(root1 == null&&root2 != null)return false;
        if(root2 == null)return true;
        if(root1.val != root2.val)return false;
        return IsSubtree(root1.left,root2.left)&&IsSubtree(root1.right,root2.right);
    }
}</span>

方法二：

public class IdenticalTree {
    public boolean chkIdentical(TreeNode A, TreeNode B) {
        // write code here
        public boolean chkIdentical(TreeNode A, TreeNode B) {
        // write code here
        String str1=treeToStr(A);
        String str2=treeToStr(B);
         
        int res=getIndexOf(str1,str2);
         
        if(res==-1)
        {
            return false;
        }else{
            return true;
        }
    }
     
     
    public String treeToStr(TreeNode tree)
    {
        if(tree==null)
        {
            return "#";
        }
         
       StringBuilder sb=new StringBuilder();
       sb.append(tree.val);
       sb.append(treeToStr(tree.left));
       sb.append(treeToStr(tree.right));
       return sb.toString();
    }
     
    public int getIndexOf(String s1,String s2)
    {
        if(s1==null||s2==null||s1.equals("")||s2.equals(""))
        {
            return -1;
        }
         
        int[] next=getNext(s2);
         
        char[] sc1=s1.toCharArray();
        char[] sc2=s2.toCharArray();
         
        int m1=0;
        int m2=0;
         
        while(m1<sc1.length&&m2<sc2.length)
        {
            if(sc1[m1]==sc2[m2])
            {
                m1++;
                m2++;
            }else if(next[m2]==-1)
            {
                m1++;
            }else{
                m2=next[m2];
            }
        }
         
        return m2==sc2.length?m1-m2:-1;
    }
     
    public int[] getNext(String str)
    {
        int[] next=new int[str.length()];
        char[] s=str.toCharArray();
        next[0]=-1;
        next[1]=0;
         
        if(s.length<2)
        {
            return next;
        }
         
        int cur=2;
        int cn=0;
        while(cur<s.length)
        {
            if(s[cur-1]==s[cn])
            {
                next[cur++]=++cn;
            }else if(cn>0)
            {
                cn=next[cn];
            }else{
                next[cur++]=0;
            }
        }
        return next;
         
    }
    }
}