1020 Tree Traversals
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 7 • 1 • 2 • 3
Sample Output:
4 1 6 3 5 7 2
题意
第一行包含整数 N ，表示二叉树的节点数。
第二行包含 N 个整数，表示二叉树的后序遍历。
第三行包含 N 个整数，表示二叉树的中序遍历。
输出一行 N 个整数，表示二叉树的层序遍历。
思路
- 由于题目给定每个结点的值都不同，所以可以用哈希表存储每个结点在中序数组中的下标，方便后续使用。
- 通过下标构建二叉树（其中
k-1-il表示左子树结点个数）：
- 通过
bfs进行层次遍历，并输出最终结果。
代码
#include<bits/stdc++.h> using namespace std; const int N = 40; int n; int posorder[N], inorder[N]; unordered_map<int, int> l, r, pos; int q[N]; //构建树 int build(int il, int ir, int pl, int pr) { int root = posorder[pr]; //获得根结点 int k = pos[root]; //获得该结点在中序数组中的下标 if (il < k) l[root] = build(il, k - 1, pl, pl + (k - 1 - il)); //递归左子树 if (k < ir) r[root] = build(k + 1, ir, pl + (k - 1 - il) + 1, pr - 1); //递归右子树 return root; } //层次遍历 void bfs(int root) { //初始化，用队列来模拟层次遍历 int hh = 0, tt = 0; q[0] = root; //开始遍历 while (hh <= tt) { int k = q[hh++]; if (l.count(k)) q[++tt] = l[k]; if (r.count(k)) q[++tt] = r[k]; } //输出结果 cout << q[0]; for (int i = 1; i < n; i++) cout << " " << q[i]; cout << endl; } int main() { cin >> n; for (int i = 0; i < n; i++) cin >> posorder[i]; for (int i = 0; i < n; i++) { cin >> inorder[i]; pos[inorder[i]] = i; //记录该值在中序数组中的下标 } int root = build(0, n - 1, 0, n - 1); bfs(root); return 0; }
