Sword finger offer 26 Substructure of tree

Input two binary trees A and B, Judge B Is it right? A Substructure of .( A convention empty tree is not a substructure of any tree )

B yes A Substructure of , namely A There are emergence and B Same structure and node values .

for example :
Given tree A:

     3
    / \
   4   5
  / \
 1   2

Given tree B：

   4 
  /
 1

return true, because B And A A subtree of has the same structure and node values .
Example 1：

Input ：A = [1,2,3], B = [3,1]
Output ：false
Example 2：

Input ：A = [3,4,5,1,2], B = [4,1]
Output ：true

class Solution {
    
    public boolean isSubStructure(TreeNode A, TreeNode B) {
      // Judge by ab A tree for roots  b Whether it is a Substructure of 
    // The function of this function is to judge the two trees of the input parameter ,b Whether it is a Substructure of 
    // The principle is to see a Whether or not b Exactly the same , If the same returns true, If not, judge a Whether the left and right subtrees of are related to b Exactly the same ,
    // The principle is to find A A child node of the tree （ Contains the root node ） Is the subtree of the root and B The trees are exactly equal , If this condition is met , It can be said that it meets the meaning of the topic 
    // If traversal to empty has not found , It proves that the meaning of the question is not satisfied 
        if (B == null || A == null) {
      // Special value judgment （ Recursion to empty has not been found through the following if）
            return false;
        } 

        if (A.val == B.val && (helper(A.left, B.left) && helper(A.right, B.right))) {
      // Judge ab Is it completely equal 
            return true;    // There is only one recursive function true return , So the top function is true（ Because recursion uses the or operator ）
        }
        
        
        // Recursively call （ If the above if If you pass, you will not enter here. Enter here to explain ab A tree with roots does not meet the condition ）
        // If the above if Pass to prove B Trees and A The molecular structure of a part of the tree is completely equal  
        return isSubStructure(A.left, B) || isSubStructure(A.right, B);  

        // Use the or operator , As long as there is one satisfaction, it will be satisfied as a whole 
        
       
      
    }

    private boolean helper(TreeNode root1, TreeNode root2) {
      // Judge by ab Whether the trees with roots are completely equal 
    // Be careful , The function of this function is to judge whether the two trees of the input parameter are completely equal 
        if (root2 == null) {
    
            return true;
        }
        if (root1 == null) {
    
            return false;
        }
        if (root1.val == root2.val) {
    
            return helper(root1.left, root2.left) && helper(root1.right, root2.right);
        } else {
    
            return false;
        }
    }
}