1020. Tree Traversals (25)
2015-06-23 16:47
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题目链接:http://www.patest.cn/contests/pat-a-practise/1020
题目:
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:
Sample Output:
分析:
题目要求的是根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:
截图:
——Apie陈小旭
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:
题目要求的是根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++){//由后续遍历找到根节点,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(如果有)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\n", tmp -> c); } else printf("%d ", tmp -> c); } } return 0; }
截图:
——Apie陈小旭
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