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# 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

### 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* 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);
cout << ans.front();
ans.pop();
while(!ans.empty()) {
cout << ' ' << ans.front();
ans.pop();
}
cout << endl;
return 0;
}
 
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• 原文地址：https://www.cnblogs.com/bianjunting/p/12555671.html