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

思路：由中序序列和后续序列构造一棵树我想不用多说了，用递归构造(后序遍历的最后一个节点是树根节点，确定这个根节点R，在中序遍历中，R的左边是它的左子树，右边是它的右子树，再递归)。然后这题的重点是“蛇形风骚遍历”，我是这样想的，规定一个vector v，定义int last指向本层的最后一个节点，首先将root节点放进vector，读完root后，令last = v.size();(此时是1，也就是第二层的第一个节点位置)，然后将root节点的左右孩子放入v(如果不为NULL的话)，下一层应该是从last到v.size()这些节点，定义一个flag，每读完一层让flag++，利用flag的奇偶性决定应该从last读到v.size()还是从v.size()读到last。这一层读完之后，再另last = v.size()也就是下一层的开始，同时把本层的各个节点的孩子节点放入v中...以此类推，知道v.size() = n为止。代码：

#include<bits/stdc++.h>#define MAX 31using namespace std;class Node{public:    int data;    Node *l = NULL;    Node *r = NULL;};int n;int inOrder[MAX];int postOrder[MAX];void zigzaggingOrder(Node* root){    vector<Node*> v;    v.push_back(root);    int flag = 0;    int last = 0;    while(true){        if(v.size() == 1)            cout<<v[0]->data;        else{            if(flag % 2 == 1)                for(int i = v.size() - 1; i >= last; i--)                    cout<<" "<<v[i]->data;            else                for(int i = last; i < v.size(); i++)                    cout<<" "<<v[i]->data;            flag++;        }        if(v.size() == n)break;        int temp = v.size();        for(int i = last; i < temp; i++){            if(v[i]->l != NULL) v.push_back(v[i]->l);            if(v[i]->r != NULL) v.push_back(v[i]->r);        }        last = temp;    }    cout<<endl;}//由中序序列和后续序列构造一棵树Node* create(int ib,int ie,int pb,int pe){    if(ib > ie)        return NULL;    Node* node = new Node();    node -> data = postOrder[pe];    int ibt = ib;    while(inOrder[ibt] != postOrder[pe]) ibt++;        node -> l = create(ib,ibt - 1,pb,pb + ibt - ib - 1);        node -> r = create(ibt + 1,ie,pb + ibt - ib,pe - 1);    return node;}int main(){    cin>>n;    for(int i = 0; i < n; i++)        cin>>inOrder[i];    for(int i = 0; i < n; i++)        cin>>postOrder[i];    Node* root = create(0,n - 1,0 ,n - 1);    zigzaggingOrder(root);}