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
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?' ':' '); 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; }