1127. ZigZagging on a Tree (30)

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. A unique binary tree can be determined by a given pair of postorder and inorder traversal sequences. And it is a simple standard routine to print the numbers in level-order. However, if you think the problem is too simple, then you are too naive. This time you are supposed to print the numbers in "zigzagging order" -- that is, starting from the root, print the numbers level-by-level, alternating between left to right and right to left. For example, for the following tree you must output: 1 11 5 8 17 12 20 15.

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 inorder sequence and the third line gives the postorder sequence. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print the zigzagging sequence of the tree in a line. 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:

812 11 20 17 1 15 8 512 20 17 11 15 8 5 1

Sample Output:

1 11 5 8 17 12 20 15

建树，分奇偶输出即可

#include<stdio.h>#include<stdlib.h>#include<queue>#include<vector>using namespace std;typedef struct Node *node;struct Node{int x;int depth;node left;node right;};int in[35],post[35];vector<int>level[35];node build(int s1,int e1,int s2,int e2){node T=(node)malloc(sizeof(struct Node));T->x=post[e2];T->left=T->right=NULL;int i,root;for(i=s1;i<=e1;i++){if(in[i]==post[e2]){root=i;break;}}if(root!=s1){T->left=build(s1,root-1,s2,root-1-s1+s2);}if(root!=e1){T->right=build(root+1,e1,root-1-s1+s2+1,e2-1);}return T;}int main(){int n,i,j;scanf("%d",&n);for(i=0;i<n;i++){scanf("%d",&in[i]);}for(i=0;i<n;i++){scanf("%d",&post[i]);}node T=build(0,n-1,0,n-1);queue<node>q;T->depth=1;q.push(T);int deepest=0;while(!q.empty()){node head=q.front();q.pop();if(head->depth>deepest){deepest=head->depth;}level[head->depth].push_back(head->x);if(head->left){head->left->depth=head->depth+1;q.push(head->left);}if(head->right){head->right->depth=head->depth+1;q.push(head->right);}}printf("%d",T->x);for(i=2;i<=deepest;i++){if(i%2==0){for(j=0;j<level[i].size();j++){printf(" %d",level[i][j]);}}else{for(j=level[i].size()-1;j>=0;j--){printf(" %d",level[i][j]);}}}}