algorithm - "cracking the coding interview(fourth edition)": 4. 7 子树检查

Question

You have two very large trees: T1, with millions of nodes, and T2, with hundreds of nodes. Create an algorithm to decide if T2 is a subtree of T1

作者给出了一个暴力搜索的解决方案，做一下对比一个节点一个节点。以下是代码:

boolean containsTree(TreeNode t1, TreeNode t2) {
    if (t2 == null) return true;
    else return subTree(t1, t2);
}

boolean subTree(TreeNode r1, TreeNode r2) {
    if(r1 == null)
        return false;
    if(r1.data = r2.data){
        if(matchTreee(r1, r2)) return true;
    }
    return ( subTree(r1.left, r2) || subTree(r1.right, r2) );
}

boolean matchTree(TreeNode r1, TreeNode r2){
    if (r2 == null && r1 == null)
        return true;
    if (r1 == null || r2 == null)
        return false; //big tree empty & subtree still not found
    if (r1.data != r2.data)
        return false;
    return (matchTree(r1.left, r2.left) && matchTree(r1.right, r2.right));
}

在上面的代码中，我不同意函数 matchTree 的基本情况

if (r2 == null && r1 == null)
    return true;
if (r1 == null || r2 == null)
    return false; // big tree empty & subtree still not found

根据我的理解，基本情况应该是:

    if(r2 == null) return true; //if r2 is null, r2 must match r1 
no matter r1 is null or not
    if(r1 == null) return false; //r1 == null means big tree empty 
and we already know r2 != null, so r2 must not match r1.

大家可以帮我验证一下吗？

谢谢，

Answer 1

在我看来，这本书的作者对 T2 is a subtree of T1 的定义与您的不同。

例如，考虑下面的树(疯狂的 Paint 技能):

作者的代码认为 T3 是 T1 的子树，而不是 T2。你认为 T2 和 T3 都是 T1 的子树。

也就是说，我认为您的基本案例对于“子树”的“自然”定义更有意义。

C# 测试代码(带有int 类型的data):

var t1 = new TreeNode() { data = 0,
    left = new TreeNode() { data = 1,
        left = new TreeNode() { data = 2 },
        right = new TreeNode() { data = 3, 
            right = new TreeNode() { data = 4 } } } };

var t2 = new TreeNode() { data = 1,
    left = new TreeNode() { data = 2 },
    right = new TreeNode() { data = 3 } };

var t3 = new TreeNode() { data = 1,
    left = new TreeNode() { data = 2 },
    right = new TreeNode() { data = 3,
        right = new TreeNode() { data = 4 } } };