• 1020. Tree Traversals (25)


    题目链接:http://www.patest.cn/contests/pat-a-practise/1020

    题目:

    1020. Tree Traversals (25)

    时间限制
    400 ms
    内存限制
    65536 kB
    代码长度限制
    16000 B
    判题程序
    Standard
    作者
    CHEN, Yue

    Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.

    Input Specification:

    Each input file contains one test case. For each case, the first line gives a positive integer N (<=30), the total number of nodes in the binary tree. The second line gives the postorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.

    Output Specification:

    For each test case, print in one line the level order traversal sequence of the corresponding binary tree. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.

    Sample Input:
    7
    2 3 1 5 7 6 4
    1 2 3 4 5 6 7
    
    Sample Output:
    4 1 6 3 5 7 2

    分析:

    题目要求的是依据树的后序遍历和中序遍历构造树,再层序遍历输出。这就考察我们三点:

    1)树的结构体会不会,新节点的生成等等。

    2)两种遍历构造树会不会,用递归函数写

    3)层序遍历会不会,用队列辅助(层序、广搜用队列;深搜用递归、栈)

    AC代码:

    #include<stdio.h>
    #include<queue>
    using namespace std;
    struct Node{//节点结构体
     Node *lchild;
     Node*rchild;
     int c;
    }Tree[32];
    int loc;
    int str1[32], str2[32];//分别存储后序遍历和中序遍历的字符串
    Node *create(){
     Tree[loc].lchild = Tree[loc].rchild = NULL;
     return &Tree[loc++];
    }
    queue<Node *>Q;//用于层序遍历,用队列
    Node *build(int s1, int e1, int s2, int e2){//利用递归来构建数
     Node *ret = create();
     ret->c = str1[e1];
     int rootIdx;
     for (int i = s2; i <= e2; i++){//由兴许遍历找到根节点,然后找到其在中序的位置
      if (str2[i] == str1[e1]){
       rootIdx = i;
       break;
      }
     }
     if (rootIdx != s2){//然后进行左右子树的分隔
      ret->lchild = build(s1, s1 + rootIdx - s2 - 1, s2, rootIdx - 1);
     }
     if (rootIdx != e2){
      ret->rchild = build(s1 + rootIdx - s2, e1 - 1, rootIdx + 1, e2);
     }
     return ret;
    }
    int main(void){
     //freopen("F://Temp/input.txt", "r", stdin);
     int n;
     while (scanf("%d", &n) != EOF){
      for (int i = 0; i < n; i++){
       scanf("%d", &str1[i]);
      }
      for (int i = 0; i < n; i++){
       scanf("%d", &str2[i]);
      }
      loc = 0;
      Node *T = build(0, n - 1, 0, n - 1);
      Q.push(T);
      while (!Q.empty()){//用队列来层序遍历输出。把对头元素取出输出,并让左右节点(假设有)入队
       Node *tmp = Q.front();
       if (tmp -> lchild != NULL){
        Q.push(tmp->lchild);
       }
       if (tmp -> rchild != NULL){
        Q.push(tmp -> rchild);
       }
       Q.pop();
       if (Q.empty()){
        printf("%d
    ", tmp -> c);
       }
       else printf("%d ", tmp -> c);
      }
     }
     return 0;
    }


    截图:

    ——Apie陈小旭

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  • 原文地址:https://www.cnblogs.com/blfshiye/p/5125261.html
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