标签:ccf printf for sans oid sed return traversal UNC
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.
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.
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.
7
2 3 1 5 7 6 4
1 2 3 4 5 6 7
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
标签:ccf printf for sans oid sed return traversal UNC
原文地址:https://www.cnblogs.com/zlrrrr/p/10194858.html