Question

我试图以更易读的方式打印二叉树，而不是仅仅排成一行。我使用this question的答案作为开头，但是从左到右打印数据如下：

我需要它看起来像这样：

        25
    15       30
10    20        35

这是我的代码：

void printTree(AVLNode* root, int indent)
{
    if (root != nullptr) {
        if (indent) {
            cout << setw(indent) << ' ';
        }
        cout << root->data << endl;
        if (root->left) printTree(root->left, indent + 4);
        if (root->right) printTree(root->right, indent + 4);
    }
}

关于如何让它以我想要的方式打印的任何想法？

Answer 1

考虑以下示例

#include <iostream>
#include <string>
#include <iomanip>
using namespace std;

// .... your code ....

void buildTree(AVLNode* root, int scrWidth, int itemWidth)
// breadth-first traversal with depth limit based on screen width and output field width for one elemet
{
    bool notFinished = false;
    // check the root
    if (root)
    {
        notFinished = true;
    }
    // calculate maximum possible depth
    int depth = 1;
    int field = scrWidth;
    while (field > itemWidth)
    {
        depth++;
        field /= 2;
    }
    // check result
    if (depth < 1)
    {
        cout << " -= erroneous output options =-" << endl;
        return;
    }
    AVLNode** pItems = new AVLNode*[1];
    *pItems = root; // pointer to item on the first level
    int itemCnt = 1;
    int divWidth = 1;
    // loop for output not more than depth levels until the data is not finished
    // where level is current depth of tree, and root is on the first level
    for (int level = 1; level <= depth && notFinished; level++)
    {
        itemCnt = (level == 1) ? 1 : (itemCnt * 2);
        divWidth *= 2;
        // make list of pointers to refer items on next level
        AVLNode** list = new AVLNode*[itemCnt * 2];
        // output all utems of that level
        int nextCnt = 0;
        notFinished = false;
        for (int i = 0; i < itemCnt; i++, nextCnt += 2)
        {
            int curWidth = (scrWidth / divWidth) * ((i > 0) ? 2 : 1);
            cout << setw((curWidth>=itemWidth) ? curWidth:(itemWidth/(1+(i==0))));
            if (pItems[i])
            {
                cout << pItems[i]->data;
                list[nextCnt] = pItems[i]->left;
                list[nextCnt + 1] = pItems[i]->right;
                if (list[nextCnt] || list[nextCnt + 1])
                    notFinished = true;
            }
            else
            {
                cout << ".";
                list[nextCnt] = NULL;
                list[nextCnt + 1] = NULL;
            }
        }
        cout << endl;
        // free the memory allocated for list of pointers
        if (pItems)
            delete[] pItems;
        pItems = list; // and shift to new one (for next level)
    }
    delete[] pItems;
}

int main(int argc, char* argv[])
{
    // create some structure
    AVLNode * root = NULL;
    // code for making tree
    // ....
    buildTree(root, 80, 5);
    // some other code
    // ....
    return 0;
}

调用buildTree(root, 80, 5);打印树状（其中.表示NULL而不是item）：

                                     64
                  58                                       .
        24                  62                   .                   .
   0        78         .         .         .         .         .         .
41   69    .    .    .    .    .    .    .    .    .    .    .    .    .    .

但buildTree(root, 40, 10);会输出相同的数据

               64
     58                   .
24        62         .         .

即。只有三层，因为第四层有8个项目，如果每个要求10个字符总重量40是不够的。

注意：我没有足够的时间来调试代码并使其完美，但我希望它能帮助您找到自己的解决方案。

打印二进制树 - C ++

1 个答案: