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对称的二叉树【树的遍历】
2022-07-07 01:25:00 【MC快乐的苦Xiao怕】
题
思路:dfs遍历n个节点,判断是否对称:
#include <bits/stdc++.h>
using namespace std ;
const int N = 10000010 ;
int l[N] , r[N] , a[N] ;
int m , ans = 1 ;
bool flag ;
inline int dfs (int x , int y , int s)
{
if (x == -1 && y == -1) return 0 ;
if (x == -1 || y == -1 && x != y)
{
flag = true ;
return 0 ;
}
if (a[x] != a[y])
{
flag = true ;
return 0 ;
}
return dfs (l[x] , r[y] , 2) + dfs (r[x] , l[y] , 2) + s ;
}
int main ()
{
scanf ("%d" , & m) ;
for (int i = 1; i <= m; i++) scanf ("%d" , & a[i]) ;
for (int i = 1; i <= m; i++) scanf ("%d%d" , & l[i] , & r[i]) ;
for (int i = 1; i <= m; i++)
{
int s = dfs (l[i] , r[i] , 3) ;
if (s > ans && flag == false) ans = s ;
flag = false ;
}
printf ("%d" , ans) ;
return 0 ;
}
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