树的子结构

题目链接:https://www.nowcoder.com/practice/6e196c44c7004d15b1610b9afca8bd88?tpId=13&tqId=11170&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking

 

/**
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

    public TreeNode(int val) {
        this.val = val;

    }

}
*/
public class Solution {
      public boolean HasSubtree(TreeNode root1,TreeNode root2) {
        if(root1 == null || root2 == null){
            return false;
        }
        //判断是否为子结构，分别进行讨论比较
        return isSubTree(root1,root2)||
               isSubTree(root1.left,root2)||isSubTree(root1.right,root2);
    }
    
    public boolean isSubTree(TreeNode root1,TreeNode root2){
        //最先需要判断root2,如果此时root2为空，则表示已经比较完毕，并且root2不为空，事先已知
         if(root2 == null){
          return true;  
         }
         if(root1 == null){
             return false;
         }
         if(root1.val != root2.val){
             return false;
         }
        //此时可知，root1.val 和 root2.val相等，接下来再比较其余部分
        return isSubTree(root1.left,root2.left) && isSubTree(root1.right,root2.right);
    }
    
}