C++ 为二叉树取多个输入

Question

I am struggling how to handle multiple inputs for the binary tree.

The input format is:

• The first line takes an integer n as an input representing level of tree.
• Next n + 1 lines contain the nodes in the tree in level order.

Example input:

2
1
2 3
4 5 6 7

Pictorial representation of above inputs:

     1
    / \
   /   \
  2     3
 / \   / \
4   5 6   7

I want to put those values from second line to last line into vector, but it is hard to get the inputs like above.

Here is what I tried:

int main(void) {
    vector<int> tree;
    int lv, num;
    int treeSize = 1;

    cin >> lv;

    for(int i = 0; i <= lv; i++) {
        while(tree.size() < treeSize && cin >> num) {
            tree.push_back(num);
        }
        treeSize += (i + 1) * 2;
    }

    return 0;
}

My code doesn't work properly. For instance, above example input has to have total 7 nodes, but if I put too many numbers in one line as input, it exceeds 7 nodes.

Is there any proper way to get inputs in case of this?

Answer 1

The code looks correct at a glance. I would trace it in a debugger. IMHO your code is a bit difficult. I would do the array filling like below.

int main(void) {
    vector<int> tree;
    int lv, num;

    cin >> lv;

    for(int i = 0, lvSize = 1; i <= lv; i++; lvSize *= 2) {
        for (int j = 0; j < lvSize; && cin >> num; j++) {
            tree.push_back(num);
        }
    }

    return 0;
}

Since you read a tree in a vector, you know the number of elements and the code can be simplified and reduced a bit more.

int main(void) {
    vector<int> tree;
    int lv, num;

    cin >> lv;
    int size = (1 << (lv + 1)) - 1;   // This is pow(2, lv + 1) - 1

    for(int i = 0; i < size; i++) {
        tree.push_back(num);
    }

    return 0;
}