链接

树的同构

题意

同构的定义: 给定两棵树T1和T2。如果T1可以通过若干次左右孩子互换就变成T2，则我们称两棵树是“同构”的。

给两棵树，判断是否同构

做法

先建树，然后判断，知道怎么判断就知道怎么写了，具体实现看代码

代码

/*    Name: hello world.cpp    Author: AA    Description: 唯代码与你不可辜负*/#include<bits/stdc++.h>using namespace std;#define LL long long#define maxn 20100typedef struct Node {    char data;    int left;    int right;} Tree;Tree t1[maxn], t2[maxn];int build(Tree  T[]) {    int n, cherk[maxn], root = -1;    char cl, cr;    scanf("%d", &n);    getchar();    if(n) {        for(int i = 0; i < n; i++)            cherk[i] = 0;        for(int i = 0; i < n; i++) {            scanf("%c %c %c", &T[i].data, &cl, &cr);            getchar();            if(cl != ‘-‘) {                T[i].left = cl - ‘0‘;                cherk[T[i].left] = 1;            } else                T[i].left = -1;            if(cr != ‘-‘) {                T[i].right = cr - ‘0‘;                cherk[T[i].right] = 1;            } else                T[i].right = -1;        }        for(int i = 0; i < n; i++) {            if(cherk[i] == 0) {                root = i;                break;            }        }    }    return root;}bool Isomorphic(int root1, int root2) {    if(root1 == -1 && root2 == -1) return true;//都是空的，符合    else if((root1 == -1 && root2 != -1) || (root1 != -1 && root2 == -1)) return false; //一个空一个不空当然不符合    else if(t1[root1].data != t2[root2].data) return false; //元素不相等不符合    else if(t1[root1].left == -1 && t2[root2].left == -1) //左子树都是空的话,就判断右子树        return Isomorphic(t1[root1].right, t2[root2].right);    else if(t1[t1[root1].left].data =http://www.mamicode.com/= t2[t2[root2].left].data)//左子树的元素都相同的话就把他们的左右子树都判断        return (Isomorphic(t1[root1].left, t2[root2].left) && Isomorphic(t1[root1].right, t2[root2].right));    else //最后是把左右子树交换来判断        return (Isomorphic(t1[root1].left, t2[root2].right) && Isomorphic(t1[root1].right, t2[root2].left));}int main() {    int root1, root2;    root1 = build(t1);    root2 = build(t2);    if(Isomorphic(root1, root2))        puts("Yes");    else        puts("No");}