pat 1020. Tree Traversals (25)

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:

解：根据后序和中序遍历的结果，确定二叉树的结构，再分层遍历。

代码：

#include <iostream>
#include <fstream>
#include <string>
#include <vector>
using namespace std;
struct TreeNode
{
    struct TreeNode* left;
    struct TreeNode* right;
    int  elem;
};
TreeNode* BinaryTreeFromOrderings(int* inorder, int* aftorder, int length)
{
    if(length == 0)
    {
        return NULL;
    }
    TreeNode* node = new TreeNode;//Noice that [new] should be written out.
    node->elem = *(aftorder+length-1);
    int rootIndex = 0;
    for(;rootIndex < length; rootIndex++)//a variation of the loop
    {
        if(inorder[rootIndex] ==  *(aftorder+length-1))
            break;
    }
    node->left = BinaryTreeFromOrderings(inorder, aftorder , rootIndex);
    node->right = BinaryTreeFromOrderings(inorder + rootIndex + 1, aftorder + rootIndex , length - (rootIndex + 1));
    return node;
}

void PrintNodeByLevel(TreeNode*pRoot)
{
       if(pRoot==NULL)
              return;
       vector<TreeNode*>vec;
       //运用了vector可以利用其动态增长的特性
       //但是这个时候标号就不能用迭代器了，因为增长的时候迭代器会失效
       vec.push_back(pRoot);
       int intcur=0;//指向当前节点
       int intlast=1,cur=0,last,flag=0;
       while(cur<vec.size())
       {
              last=vec.size();//指向本层节点的后一个节点
              if(vec[cur]->elem!=0&&flag==1)
              {
                  printf(" %d",vec[cur]->elem);
              }
              if(vec[cur]->elem!=0&&flag==0)
              {
                  flag=1;
                  printf("%d",vec[cur]->elem);
              }
              if(vec[cur]->left!=NULL)
                     vec.push_back(vec[cur]->left);
              if(vec[cur]->right!=NULL)
                     vec.push_back(vec[cur]->right);
              cur++;
       }
       //每一次循环都将下一层的节点压入vector，并且打印完本层节点
}
int main(int argc, char** argv)
{
    int n;
    while(scanf("%d",&n)==1)
    {
        int *af=new int[n+1];
        int *in=new int[n+1];
        for(int i=0;i<n;i++)
        {
            scanf("%d",&af[i]);
        }
        for(int i=0;i<n;i++)
        {
            scanf("%d",&in[i]);
        }
        PrintNodeByLevel(BinaryTreeFromOrderings(in, af, n));
        printf("
");
    }
    return 0;
}

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