PAT1020: Tree Traversals
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1020. Tree Traversals (25)
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 7Sample Output:
4 1 6 3 5 7 2
思路
1.由后序遍历序列的最后一位确定树的根节点,根据根节点在中序遍历序列的位置确定左右子树,然后按照这个规律不停地分割两个序列就行,
2.分割以前序遍历的形式进行,用一个二维vector levels存储遍历到的节点levels[level][i]和其对应层级level,最后输出levels不为空的部分就是层次遍历。
3.输出需要注意最后一个节点输出后面不能有空格
代码
#include<iostream> #include<vector> using namespace std; vector<vector<int>> levels(31); vector<int> postorder(31); vector<int> inorder(31); int index; void findroot(int pfirst,int plast,int ifirst,int ilast,int level) { if(ifirst > ilast || pfirst > plast) return; int i = 0; while(postorder[plast] != inorder[i+ifirst]) i++; levels[level].push_back(postorder[plast]); findroot(pfirst,pfirst + i - 1,ifirst,ifirst + i - 1,level + 1); findroot(pfirst + i,plast - 1,ifirst + i + 1,ilast,level + 1); } int main() { int N; while(cin >> N) { for(int i = 1;i <= N;i++) { cin >> postorder[i]; } for(int i = 1;i <= N;i++) { cin >> inorder[i]; } index = 0; //preorder and store every level‘s node findroot(1,N,1,N,1); //print int lastout = 0; //to ensure the last output has no space in the end for(int i = 1;i <= N;i++) { if(levels[i].empty()) continue; for(int j = 0;j < levels[i].size();j++) { if(lastout == N - 1) cout << levels[i][j]; else { cout << levels[i][j]<<" "; ++lastout; } } } cout << endl; } }
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