PAT 甲级 1020 Tree Traversals
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https://pintia.cn/problem-sets/994805342720868352/problems/994805485033603072
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 (≤), 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:
4 1 6 3 5 7 2
代码:
#include <bits/stdc++.h> using namespace std; int N; vector<int> in, post, level(100000, -1); void rec(int root, int st, int en, int index) { if(st > en) return; int i = st; while(i < en && in[i] != post[root]) i ++; level[index] = post[root]; rec(root - 1 - en + i, st, i - 1, 2 * index + 1); rec(root - 1, i + 1, en, 2 * index + 2); } int main() { scanf("%d", &N); in.resize(N), post.resize(N); for(int i = 0; i < N; i ++) scanf("%d", &post[i]); for(int i = 0; i < N; i ++) scanf("%d", &in[i]); rec(N - 1, 0, N - 1, 0); int cnt = 0; for(int i = 0; i < level.size(); i ++) { if(level[i] != -1) { if(cnt) printf(" "); printf("%d", level[i]); cnt ++; } if(cnt == N) break; } return 0; }
已知后序中序求层序遍历的结果
vector<int> level(100000, -1); 这句话很有必要
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