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.
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
#include <iostream> #include <algorithm> #include <queue> using namespace std; struct _root { int data; _root *left, *right; }; int n; const int maxn = 30 + 5; int post[maxn], inord[maxn]; _root *head = new _root; _root* build(int pl, int pr, int il, int ir) { _root* root = new _root; root -> data = post[pr]; if(root -> data == post[n - 1]) head = root; root -> left = root -> right = NULL; int temp = find(inord + il, inord + ir + 1, post[pr]) - inord - il; if(temp > 0) root -> left = build(pl, pl + temp - 1, il, il + temp - 1); if(pl + temp <= pr - 1) root -> right = build(pl + temp, pr - 1, il + temp + 1, ir); return root; } queue <_root*> que; queue <int> ans; void print(_root *root) { if(root == NULL) return; que.push(root); while(!que.empty()) { _root * now = que.front(); que.pop(); ans.push(now -> data); if(now -> left) que.push(now -> left); if(now -> right) que.push(now -> right); } } int main() { cin >> n; for(int i = 0; i < n; i ++) cin >> post[i]; for(int i = 0; i < n; i ++) cin >> inord[i]; build(0, n - 1, 0, n - 1); print(head); cout << ans.front(); ans.pop(); while(!ans.empty()) { cout << ' ' << ans.front(); ans.pop(); } cout << endl; return 0; }