判断t1树中是否有与t2树拓扑结构完全相同的子树

最新推荐文章于 2024-02-13 05:00:54 发布

原创最新推荐文章于 2024-02-13 05:00:54 发布 · 788 阅读

0 ·

CC 4.0 BY-SA版权

本文介绍了一种通过序列化树结构来判断一棵树是否为另一棵树的子树的方法。使用前序遍历序列化树，并借助KMP算法快速查找子树是否存在。此方法巧妙地将树的结构问题转化为字符串匹配问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

/**
 * Created by lxw, liwei4939@126.com on 2017/10/30.
 * 判断t1树中是否有与t2树拓扑结构完全相同的子树
 */
public class IsContainSubTree {

    public class Node{
        public int value;
        public Node left;
        public Node right;

        public Node(int data){
            this.value = data;
        }
    }
    
    public boolean isSubTree(Node t1, Node t2){
        String t1Str = serialByPre(t1);
        String t2Str = serialByPre(t2);
        return getIndexOf(t1Str, t2Str) != -1;
    }
    
    public String serialByPre(Node head){
        if(head == null){
            return "#!";
        }
        String res = head.value + "!";
        res += serialByPre(head.left);
        res += serialByPre(head.right);
        return res;
    }
    
    // KMP
    public int getIndexOf(String s, String m){
        if(s == null || m == null || m.length() < 1 || s.length() < m.length()){
            return -1;
        }
        char[] ss = s.toCharArray();
        char[] ms = m.toCharArray();
        int si = 0;
        int mi = 0;
        int[] next = getNextArray(ms);
        while (si < ss.length && mi < ms.length){
            if(ss[si] == ms[mi]){
                si++;
                mi++;
            } else if(next[mi] == -1){
                si++;
            } else {
                mi = next[mi];
            }
        }
        return mi == ms.length ? si - mi : -1;
    }
    
    public int[] getNextArray(char[] ms){
        if(ms.length == 1){
            return new int[]{-1};
        }
        int[] next = new int[ms.length];
        next[0] = -1;
        next[1] = 0;
        int pos =2;
        int cn =0;
        while (pos< next.length){
            if(ms[pos - 1] == ms[cn]){
                next[pos++] = ++cn;
            } else if(cn > 0){
                cn = next[cn];
            } else {
                next[pos++] = 0;
            }
        }
        return next;
    }
}