题目描述
给定彼此独立的两棵二叉树，判断 t1 树是否有与 t2 树拓扑结构完全相同的子树。
设 t1 树的边集为 E1，t2 树的边集为 E2，若 E2 等于 E1 ，则表示 t1 树和t2 树的拓扑结构完全相同。
输入描述:
第一行输入两个整数 n 和 root，n 表示二叉树 t1 的总节点个数，root 表示二叉树 t1 的根节点。
以下 n 行每行三个整数 fa，lch，rch，表示 fa 的左儿子为 lch，右儿子为 rch。(如果 lch 为 0 则表示 fa 没有左儿子，rch同理)
第 n+2 行输入两个整数 m 和 root，n 表示二叉树 t2 的总节点个数，root 表示二叉树 t2 的根节点。
以下 m 行每行三个整数 fa，lch，rch，表示 fa 的左儿子为 lch，右儿子为 rch。(如果 lch 为 0 则表示 fa 没有左儿子，rch同理)
输出描述:
如果 t1 树有与 t2 树拓扑结构完全相同的子树，则输出 “true”，否则输出 “false”。
示例1
输入
9 1
1 2 3
2 4 5
4 0 8
8 0 0
5 9 0
9 0 0
3 6 7
6 0 0
7 0 0
5 2
2 4 5
4 0 8
8 0 0
5 9 0
9 0 0
输出
true
解法一：类似上一题只不过加一个判断条件
import java.io.*;
import java.util.*;
public class Main{
public static void main(String[] args) throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
br.readLine();
TreeNode r1 = createTree(br);
br.readLine();
TreeNode r2 = createTree(br);
boolean res = judge(r1,r2);
System.out.println(res);
}
public static boolean judge(TreeNode r1,TreeNode r2){
if(r1==null&&r2==null) return false;
if(r1==null||r2==null) return false;
Stack<TreeNode> s = new Stack<>();
s.push(r1);
while(!s.isEmpty()){
TreeNode p = s.pop();
if(p.val==r2.val){
if(isChild(p,r2)) return true;
}
if(p.right!=null){
s.push(p.right);
}
if(p.left!=null){
s.push(p.left);
}
}
return false;
}
public static boolean isChild(TreeNode r1,TreeNode r2){
//按照r2前序遍历
Stack<TreeNode> s1 = new Stack<>();
s1.push(r1);
Stack<TreeNode> s2 = new Stack<>();
s2.push(r2);
while(!s2.isEmpty()){
TreeNode node1 = s1.pop();
TreeNode node2 = s2.pop();
if(node1.val!=node2.val) return false;
if(node2.right!=null){
if(node1.right==null) return false;
s1.push(node1.right);
s2.push(node2.right);
}else if(node1.right!=null){
return false;
}
if(node2.left!=null){
if(node1.left==null) return false;
s1.push(node1.left);
s2.push(node2.left);
}else if(node2.left!=null){
return false;
}
}
return true;
}
//递归建树
public static TreeNode createTree(BufferedReader br){
try{
String[] ss = br.readLine().trim().split(" ");
int data = Integer.parseInt(ss[0]);
int left = Integer.parseInt(ss[1]);
int right = Integer.parseInt(ss[2]);
TreeNode root = new TreeNode(data);
if(left!=0){
root.left = createTree(br);
}
if(right!=0){
root.right = createTree(br);
}
return root;
}catch(Exception e){
return null;
}
}
}
class TreeNode{
int val;
TreeNode left;
TreeNode right;
public TreeNode(int val){
this.val = val;
}
}
解法二：同上递归
import java.io.*;
import java.util.*;
public class Main{
public static void main(String[] args) throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
br.readLine();
TreeNode r1 = createTree(br);
br.readLine();
TreeNode r2 = createTree(br);
boolean res = judge(r1,r2);
System.out.println(res);
}
public static boolean judge(TreeNode r1,TreeNode r2){
if(r1==null&&r2==null) return false;
if(r1==null||r2==null) return false;
return isChild(r1,r2)||judge(r1.left,r2)||judge(r1.right,r2);
}
public static boolean isChild(TreeNode r1,TreeNode r2){
if(r2==null&&r1==null) return true;
if(r2==null||r1==null) return false;
if(r1.val!=r2.val) return false;
return isChild(r1.left,r2.left)&&isChild(r1.right,r2.right);
}
//递归建树
public static TreeNode createTree(BufferedReader br){
try{
String[] ss = br.readLine().trim().split(" ");
int data = Integer.parseInt(ss[0]);
int left = Integer.parseInt(ss[1]);
int right = Integer.parseInt(ss[2]);
TreeNode root = new TreeNode(data);
if(left!=0){
root.left = createTree(br);
}
if(right!=0){
root.right = createTree(br);
}
return root;
}catch(Exception e){
return null;
}
}
}
class TreeNode{
int val;
TreeNode left;
TreeNode right;
public TreeNode(int val){
this.val = val;
}
}
解法二：先序序列化–>字符串匹配
import java.io.*;
import java.util.*;
public class Main{
public static void main(String[] args) throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
br.readLine();
TreeNode r1 = createTree(br);
br.readLine();
TreeNode r2 = createTree(br);
boolean res = judge(r1,r2);
System.out.println(res);
}
public static boolean judge(TreeNode r1,TreeNode r2){
String s1 = preOrder(r1);
String s2 = preOrder(r2);
//KMP匹配
return match(s1,s2);
}
public static boolean match(String s1,String s2){
if(s1==null||s2==null||s1.length()<s2.length()||s2.length()<1) return false;
char[] arr1 = s1.toCharArray();
char[] arr2 = s2.toCharArray();
int[] next = getNext(arr2);
int l1 = 0;
int l2 = 0;
while(l1<arr1.length&&l2<arr2.length){
if(arr1[l1]==arr2[l2]){
l1++;
l2++;
}else if(next[l2]>=0){
l2 = next[l2];
}else{
l1++;
}
}
return l2==arr2.length?true:false;
}
public static int[] getNext(char[] arr){
if(arr.length==1) return new int[]{-1};
int[] next = new int[arr.length];
next[0] = -1;
next[1] = 0;
int pos = 2;
int cn = 0;
while(pos<arr.length){
if(arr[pos-1]==arr[cn]){
next[pos++] = ++cn;
}else if(cn==0){
next[pos++] = 0;
}else{
cn = next[cn];
}
}
return next;
}
//前序序列化
public static String preOrder(TreeNode r){
if(r==null) return "#!";
String res = r.val+"!";
res += preOrder(r.left);
res += preOrder(r.right);
return res;
}
//递归建树
public static TreeNode createTree(BufferedReader br){
try{
String[] ss = br.readLine().trim().split(" ");
int data = Integer.parseInt(ss[0]);
int left = Integer.parseInt(ss[1]);
int right = Integer.parseInt(ss[2]);
TreeNode root = new TreeNode(data);
if(left!=0){
root.left = createTree(br);
}
if(right!=0){
root.right = createTree(br);
}
return root;
}catch(Exception e){
return null;
}
}
}
class TreeNode{
int val;
TreeNode left;
TreeNode right;
public TreeNode(int val){
this.val = val;
}
}
