1020. Tree Traversals (25)

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.

7
2 3 1 5 7 6 4
1 2 3 4 5 6 7

4 1 6 3 5 7 2

#include<iostream>
#include<queue>
using namespace std;
int a[32], b[32];
struct Node {
	int val;
	Node *left;
	Node *right;
};
Node* BuildTree(int *da,int *db,int n) {
	int i, j;
	if (n <= 0)
		return NULL;
	for (i = 0; i < n; i++) {
		if (*(db + i) == *(da + n - 1))
			break;
	}
	//important
	Node *nodeTemp = (Node*)malloc(sizeof(Node));
	nodeTemp->val = *(db + i);
	nodeTemp->left = BuildTree(da, db, i);
	nodeTemp->right = BuildTree(da + i, db + i + 1, n - i - 1);
	return nodeTemp;
}
int main(void) {
	int n;
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> a[i];
	}
	for (int i = 0; i < n; i++) {
		cin >> b[i];
	}
	Node* root = (Node*)malloc(sizeof(Node));
	root = BuildTree(a, b, n);
	queue<Node*> level;
	level.push(root);
	int flag = 0;
	while (true)
	{
		if (level.front()->left != NULL)
			level.push(level.front()->left);
		if (level.front()->right != NULL)
			level.push(level.front()->right);
		cout << level.front()->val;
		level.pop();
		if (level.empty())
			break;
		cout << " ";
	}
	return 0;
}