1020 Tree Traversals (25 分)
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
这道题我感觉像是在测你脑回路，挺有趣的，
这题我用bfs写的，先给你后序遍历（左右中）和中序遍历（左中右）。
也有看到一些别的题解，比如柳诺大神的题解，但是她的解法如果数据强点就过不去了。
所以后台数据确实比较水，但是用bfs的代码就不用担心数据量的问题了。
比如下面这组数据：
30 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
这组数据她的代码绝对跑不了，因为2^30可是超了int，就算是vector也不可能开这么多，而且那样很浪费空间
用个bfs写出来就时间空间都解决了。
1 #include <bits/stdc++.h> 2 #define N 100 3 using namespace std; 4 int an[40], bn[40], n, cn[N],pos = 0; 5 6 struct Node{ 7 int root, start, end; 8 }; 9 queue<Node> q; 10 void dfs(int root, int start, int end){ 11 Node node; 12 node.root = root, node.start = start,node.end = end; 13 q.push(node); 14 while(!q.empty()){ 15 node = q.front(); 16 q.pop(); 17 if(node.start > node.end) 18 continue; 19 cn[pos++] = an[node.root]; 20 int i; 21 for(i = node.start; i < node.end; i++){ 22 if(bn[i] == an[node.root]) 23 break; 24 } 25 q.push({node.root-(node.end-i)-1, node.start, i-1}); 26 q.push({node.root-1, i+1, node.end}); 27 } 28 } 29 30 int main(){ 31 cin >> n; 32 for(int i = 0; i < n; i++){ 33 cin >> an[i]; 34 } 35 for(int i = 0; i < n; i++){ 36 cin >> bn[i]; 37 } 38 memset(cn,-1,sizeof(cn)); 39 dfs(n-1, 0, n-1); 40 for(int i = 0; i < pos; i++) 41 printf("%d%c",cn[i],i==pos-1?‘\n‘:‘ ‘); 42 43 return 0; 44 }
这里给了一下柳神的代码（虽然是错的，但是可以过题，想法还是挺好的）：^v^(勿喷)
#include <iostream> #include <algorithm> #include <vector> using namespace std; struct node { int index, value; }; bool cmp(node a, node b) { return a.index < b.index; } vector<int> post, in; vector<node> ans; void pre(int root, int start, int end, int index) { if (start > end) return; int i = start; while (i < end && in[i] != post[root]) i++; ans.push_back({index, post[root]}); pre(root - 1 - end + i, start, i - 1, 2 * index + 1); pre(root - 1, i + 1, end, 2 * index + 2); } int main() { int n; scanf("%d", &n); post.resize(n); in.resize(n); for (int i = 0; i < n; i++) scanf("%d", &post[i]); for (int i = 0; i < n; i++) scanf("%d", &in[i]); pre(n - 1, 0, n - 1, 0); sort(ans.begin(), ans.end(), cmp); for (int i = 0; i < ans.size(); i++) { if (i != 0) cout << " "; cout << ans[i].value; } return 0; }
原文：https://www.cnblogs.com/zllwxm123/p/11033165.html