1127 ZigZagging on a Tree （30 分）

最新推荐文章于 2020-07-15 19:54:39 发布

原创最新推荐文章于 2020-07-15 19:54:39 发布 · 682 阅读

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本文介绍了一种独特的二叉树遍历方式——锯齿形层次遍历，通过给定的中序和后序遍历序列构建二叉树，并按从根节点开始的锯齿形路径输出节点值。此方法首先根据序列构建树，然后采用广度优先搜索策略，交替地从左到右和从右到左打印每一层的节点。

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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:

8
12 11 20 17 1 15 8 5
12 20 17 11 15 8 5 1

Sample Output:

1 11 5 8 17 12 20 15

#include<bits/stdc++.h>
using namespace std;
int in[1005],post[1005];
typedef struct node
{
int data;
struct node *lchild,*rchild;
}node,*Tree;
Tree Create(int inl,int inr,int postl,int postr)
{
if(postl>postr)
return NULL;
Tree root = new node;
root->data = post[postr];
int k;
for(k = inl;k<=inr;k++)
{
if(in[k]==post[postr])
break;
}
int leftnum = k-inl;
root->lchild = Create(inl,k-1,postl,postl+leftnum-1);
root->rchild = Create(k+1,inr,postl+leftnum,postr-1);
return root;
}

void bfs(Tree root)
{
queue<Tree>q;
q.push(root);
cout<<root->data;
int level=2;
while(!q.empty())
{
vector<Tree>v;
while(!q.empty())
{
Tree t= q.front();
q.pop();
if(t->lchild)
v.push_back(t->lchild);
if(t->rchild)
v.push_back(t->rchild);
}
for(int i=0;i<v.size();i++)
q.push(v[i]);
if(level%2==0)
{
for(int i=0;i<v.size();i++)
cout<<" "<<v[i]->data;
}
else
{
for(int i=v.size()-1;i>=0;i--)
cout<<" "<<v[i]->data;
}
level++;
}
}
int main()
{
int n;
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%d",&in[i]);
for(int i=0;i<n;i++)
scanf("%d",&post[i]);
Tree root;
root = Create(0,n-1,0,n-1);
bfs(root);
}