PAT-1102(Invert a Binary Tree)

　　题目见这里

　  和1099略微相似，考察二叉树和基本的遍历，算是简单的啦，下标还充当了数据域，主要是知道要标记访问到的下标,从而确定root

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//1102:Invert a Binary Tree #include <cstdio> #include <iostream>   using namespace std;   const int N = 10; typedef struct node{     int lChild,rChild; }Node;   void Input(int &root, bool *flag, Node *node, int &n){     int i;     char left,right;     scanf("%d",&n);     getchar();     for(i=0;i<n;i++){         scanf("%c %c",&left,&right);         getchar();         if(left=='-') node[i].lChild = -1;         else{             node[i].lChild = left-'0';             flag[left-'0'] = true;         }         if(right=='-') node[i].rChild = -1;         else{             node[i].rChild = right-'0';             flag[right-'0'] = true;         }     }     for(i=0;i<n;i++)         if(!flag[i]){             root = i;             break;         }  }   void Invert(int root, Node *node){     if(root!=-1){         Invert(node[root].lChild,node);         Invert(node[root].rChild,node);         swap(node[root].lChild,node[root].rChild);     } }   void LevelOrderTraverse(int root, Node *node, int n){     int qNode[N],tmpNode;     int front,rear;     front = rear = 0;     if(root==-1) return;     qNode[rear] = root;     rear ++;     while(front<rear){ //无'='         tmpNode = qNode[front];         front ++;         printf("%d",tmpNode);         if(front<n) printf(" ");         else printf("\n");         if(node[tmpNode].lChild!=-1){             qNode[rear] = node[tmpNode].lChild;             rear ++;         }         if(node[tmpNode].rChild!=-1){             qNode[rear] = node[tmpNode].rChild;             rear ++;         }     } }   void InOrderTraverse(int root, Node *node, int n){     static int i = 1;     if(root!=-1){         InOrderTraverse(node[root].lChild,node,n);         printf("%d",root);         if(i==n) printf("\n");         else printf(" ");         i ++;         InOrderTraverse(node[root].rChild,node,n);     } }   int main(){ //  freopen("Data.txt","r",stdin);     Node node[N];     bool flag[N]={false};     int root,n;     Input(root,flag,node,n);     Invert(root,node);     LevelOrderTraverse(root,node,n);     InOrderTraverse(root,node,n);     return 0; }