Sword finger offer 26. substructure of tree

Title Description

Problem analysis

This problem uses recursive method to find ：
First, judge from the root node B Is it right? A Substructure of , If not judging from the left and right subtrees respectively , Just one for true, Just explain B yes A Substructure of
If B It's empty ,A If it is not empty, the traversal ends
If B Not empty ,A It's empty , Or if the values of these two nodes are different, return false. Then compare A,B The corresponding left and right subtrees of nodes .

Code instance

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