剑指 Offer 26. 树的子结构

Posted 2021-12-06 小布丁value

力扣链接

class Solution {
    //先序遍历循环A树
      public boolean isSubStructure(TreeNode A,TreeNode B){
          return (A!=null&&B!=null)&&(recur(A,B)||isSubStructure(A.left,B)||isSubStructure(A.right,B));
       
    }
    public boolean recur(TreeNode A,TreeNode B){
        if(B==null) return true;
        if(A==null||A.val!=B.val) return false;
        return recur(A.left,B.left)&&recur(A.right,B.right);
        
    }
}