剑指 Offer 26. 树的子结构

class Solution {
    public boolean isSubStructure(TreeNode A, TreeNode B) {
        if(A==null || B==null) return false;
        if(A.val != B.val){
            return isSubStructure(A.left,B) ||isSubStructure(A.right,B);
        }else{
            return isMatched(A,B)||isSubStructure(A.left,B) ||isSubStructure(A.right,B);
        }
    }

    private boolean isMatched(TreeNode a, TreeNode b) {
        if(b!=null){
            if(a==null) return false;
            else{
                if(b.val == a.val){
                    return isMatched(a.left,b.left) && isMatched(a.right,b.right);
                }
            }
        }else{
            return true;
        }
        return false;
    }
}