Tree——No.572 Subtree of Another Tree_help him find the maximum number of consecutive da-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_41773240/article/details/88915628

本文介绍了一种使用递归方法来检查一棵树是否为另一棵树的子树的方法，并提供了一种优化方案，通过前序遍历来提高效率。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Problem：

Given two non-empty binary trees s and t, check whether tree t has exactly the same structure and node values with a subtree of s.

A subtree of s is a tree consists of a node in s and all of this node's descendants. The tree s could also be considered as a subtree of itself.

Explanation：

判断树t是否为树s的子树。

My Thinking：

使用递归，遍历每个结点，对每个结点为根的子树与s相比较。

My Solution：

class Solution {
    public boolean isSubtree(TreeNode s, TreeNode t) {
        return traverse(s,t);
    }
    public boolean traverse(TreeNode s,TreeNode t){
        if(s==null)
            return false;
        if(isEqual(s,t))
            return true;
        return traverse(s.left,t) || traverse(s.right,t);
    }
    public boolean isEqual(TreeNode s,TreeNode t){
        if((s==null && t!=null)||(s!=null && t==null))
            return false;
        if(s==null && t==null)
            return true;
        if(s.val!=t.val)
            return false;
        return isEqual(s.left,t.left) && isEqual(s.right,t.right);
    }
}

Optimum Thinking:

使用前序遍历s和t，判断t是否是s的子集，但是需要把孩子为null也考虑进去。

Optimum Solution：

class Solution {
    public boolean isSubtree(TreeNode s, TreeNode t) {
        String ts=preOrder(s,true);
        String tt=preOrder(t,true);
        return ts.indexOf(tt)>=0;
    }
    public String preOrder(TreeNode t,boolean left){
         if (t == null) {
            if (left)
                return "lnull";
            else
                return "rnull";
        }
        return "#"+t.val + " " +preOrder(t.left, true)+" " +preOrder(t.right, false);
    }
}