c++ - 使用C++以漂亮的方式打印二叉树

Question

我在尝试在c++中打印如下的二叉树时有点“迷”:

            8
           / \
          /   \
         /     \
        5       10
       / \      / \
      2   6    9   11

我知道如何获取树的高度和每个级别中的节点数，但是我无法弄清楚如何在根和第二级之间设置正确的空格数(根下有3行3个级别，但我相信并非每次都这样，我认为这可能是大树的高度的3倍)。

我想在打印行中的这些空格以及行之间的行数方面有所帮助。谢谢。

我在用C++编码

Get height

int tree::getHeight(No *node) {
  if (node == NULL) return 0;
  return 1 + max(getHeight(node->esq), getHeight(node->dir));
}

Get number of nodes per line

void tree::getLine(const No *root, int depth, vector<int>& vals){
    int placeholder = 10;
    if (depth <= 0 && root != nullptr) {
        vals.push_back(root->chave);
        return;
    }
    if (root->esq != nullptr)
       getLine(root->esq, depth-1, vals);
    else if (depth-1 <= 0)
       vals.push_back(placeholder);
    if (root->dir != nullptr)
       getLine(root->dir, depth-1, vals);
    else if (depth-1 <= 0)
       vals.push_back(placeholder);
}

Answer 1

这是创建二进制树的基于文本的表示形式的代码示例。本演示使用了最小有用的二叉树类(BinTree)，并且具有很小的占用空间，只是为了避免膨胀示例的大小。

它的文本呈现成员函数更严重，使用迭代而不是递归，如该类其他部分所述。

这分三步完成，首先将一个由字符串值组成的行 vector 放在一起。

然后，这用于格式化代表树的文本字符串行。

然后，将字符串清理干净并转交给cout。

此外，该演示还包括“随机树”功能，可连续播放数小时。

#include <iostream>
#include <vector>
#include <string>
#include <sstream>
#include <algorithm>
#include <random>

using std::vector;
using std::string;
using std::cout;

template <typename T>
class BinTree {
    struct Node {
        T value;
        Node *left,*right;
        Node() : left(nullptr),right(nullptr) {}
        Node(const T& value) :value(value),left(nullptr),right(nullptr) {}
        // stack-abusing recursion everywhere, for small code
        ~Node() { delete left; delete right; }
        int max_depth() const {
            const int left_depth = left ? left->max_depth() : 0;
            const int right_depth = right ? right->max_depth() : 0;
            return (left_depth > right_depth ? left_depth : right_depth) + 1;
        }
    };

    Node *root;

public:
    BinTree() : root(nullptr) {}
    ~BinTree() { delete root; }

    int get_max_depth() const { return root ? root->max_depth() : 0; }
    void clear() { delete root; root = nullptr; }
    void insert() {}
    template <typename ...Args>
    void insert(const T& value, Args...more) {
        if(!root) {
            root = new Node(value);
        } else {
            Node* p = root;
            for(;;) {
                if(value == p->value) return;
                Node* &pchild = value < p->value ? p->left : p->right;
                if(!pchild) { 
                    pchild = new Node(value);
                    break;
                }
                p = pchild;
            }
        }
        insert(more...);
    }

    struct cell_display {
        string   valstr;
        bool     present;
        cell_display() : present(false) {}
        cell_display(std::string valstr) : valstr(valstr), present(true) {}
    };

    using display_rows = vector< vector< cell_display > >;

    // The text tree generation code below is all iterative, to avoid stack faults.

    // get_row_display builds a vector of vectors of cell_display structs
    // each vector of cell_display structs represents one row, starting at the root
    display_rows get_row_display() const {
        // start off by traversing the tree to
        // build a vector of vectors of Node pointers
        vector<Node*> traversal_stack;
        vector< std::vector<Node*> > rows;
        if(!root) return display_rows();

        Node *p = root;
        const int max_depth = root->max_depth();
        rows.resize(max_depth);
        int depth = 0;
        for(;;) {
            // Max-depth Nodes are always a leaf or null
            // This special case blocks deeper traversal
            if(depth == max_depth-1) {
                rows[depth].push_back(p);
                if(depth == 0) break;
                --depth;
                continue;
            }

            // First visit to node?  Go to left child.
            if(traversal_stack.size() == depth) {
                rows[depth].push_back(p);
                traversal_stack.push_back(p);
                if(p) p = p->left;
                ++depth;
                continue;
            }

            // Odd child count? Go to right child.
            if(rows[depth+1].size() % 2) {
                p = traversal_stack.back();
                if(p) p = p->right;
                ++depth;
                continue;
            }

            // Time to leave if we get here

            // Exit loop if this is the root
            if(depth == 0) break;

            traversal_stack.pop_back();
            p = traversal_stack.back();
            --depth;
        }

        // Use rows of Node pointers to populate rows of cell_display structs.
        // All possible slots in the tree get a cell_display struct,
        // so if there is no actual Node at a struct's location,
        // its boolean "present" field is set to false.
        // The struct also contains a string representation of
        // its Node's value, created using a std::stringstream object.
        display_rows rows_disp;
        std::stringstream ss;
        for(const auto& row : rows) {
            rows_disp.emplace_back();
            for(Node* pn : row) {
                if(pn) {
                    ss << pn->value;
                    rows_disp.back().push_back(cell_display(ss.str()));
                    ss = std::stringstream();
                } else {
                    rows_disp.back().push_back(cell_display());
        }   }   }
        return rows_disp;
    }

    // row_formatter takes the vector of rows of cell_display structs 
    // generated by get_row_display and formats it into a test representation
    // as a vector of strings
    vector<string> row_formatter(const display_rows& rows_disp) const {
        using s_t = string::size_type;

        // First find the maximum value string length and put it in cell_width
        s_t cell_width = 0;
        for(const auto& row_disp : rows_disp) {
            for(const auto& cd : row_disp) {
                if(cd.present && cd.valstr.length() > cell_width) {
                    cell_width = cd.valstr.length();
        }   }   }

        // make sure the cell_width is an odd number
        if(cell_width % 2 == 0) ++cell_width;

        // formatted_rows will hold the results
        vector<string> formatted_rows;

        // some of these counting variables are related,
        // so its should be possible to eliminate some of them.
        s_t row_count = rows_disp.size();

        // this row's element count, a power of two
        s_t row_elem_count = 1 << (row_count-1);

        // left_pad holds the number of space charactes at the beginning of the bottom row
        s_t left_pad = 0;

        // Work from the level of maximum depth, up to the root
        // ("formatted_rows" will need to be reversed when done) 
        for(s_t r=0; r<row_count; ++r) {
            const auto& cd_row = rows_disp[row_count-r-1]; // r reverse-indexes the row
            // "space" will be the number of rows of slashes needed to get
            // from this row to the next.  It is also used to determine other
            // text offsets.
            s_t space = (s_t(1) << r) * (cell_width + 1) / 2 - 1;
            // "row" holds the line of text currently being assembled
            string row;
            // iterate over each element in this row
            for(s_t c=0; c<row_elem_count; ++c) {
                // add padding, more when this is not the leftmost element
                row += string(c ? left_pad*2+1 : left_pad, ' ');
                if(cd_row[c].present) {
                    // This position corresponds to an existing Node
                    const string& valstr = cd_row[c].valstr;
                    // Try to pad the left and right sides of the value string
                    // with the same number of spaces.  If padding requires an
                    // odd number of spaces, right-sided children get the longer
                    // padding on the right side, while left-sided children
                    // get it on the left side.
                    s_t long_padding = cell_width - valstr.length();
                    s_t short_padding = long_padding / 2;
                    long_padding -= short_padding;
                    row += string(c%2 ? short_padding : long_padding, ' ');
                    row += valstr;
                    row += string(c%2 ? long_padding : short_padding, ' ');
                } else {
                    // This position is empty, Nodeless...
                    row += string(cell_width, ' ');
                }
            }
            // A row of spaced-apart value strings is ready, add it to the result vector
            formatted_rows.push_back(row);

            // The root has been added, so this loop is finsished
            if(row_elem_count == 1) break;

            // Add rows of forward- and back- slash characters, spaced apart
            // to "connect" two rows' Node value strings.
            // The "space" variable counts the number of rows needed here.
            s_t left_space  = space + 1;
            s_t right_space = space - 1;
            for(s_t sr=0; sr<space; ++sr) {
                string row;
                for(s_t c=0; c<row_elem_count; ++c) {
                    if(c % 2 == 0) {
                        row += string(c ? left_space*2 + 1 : left_space, ' ');
                        row += cd_row[c].present ? '/' : ' ';
                        row += string(right_space + 1, ' ');
                    } else {
                        row += string(right_space, ' ');
                        row += cd_row[c].present ? '\\' : ' ';
                    }
                }
                formatted_rows.push_back(row);
                ++left_space;
                --right_space;
            }
            left_pad += space + 1;
            row_elem_count /= 2;
        }

        // Reverse the result, placing the root node at the beginning (top)
        std::reverse(formatted_rows.begin(), formatted_rows.end());

        return formatted_rows;
    }

    // Trims an equal number of space characters from
    // the beginning of each string in the vector.
    // At least one string in the vector will end up beginning
    // with no space characters.
    static void trim_rows_left(vector<string>& rows) {
        if(!rows.size()) return;
        auto min_space = rows.front().length();
        for(const auto& row : rows) {
            auto i = row.find_first_not_of(' ');
            if(i==string::npos) i = row.length();
            if(i == 0) return;
            if(i < min_space) min_space = i;
        }
        for(auto& row : rows) {
            row.erase(0, min_space);
    }   }

    // Dumps a representation of the tree to cout
    void Dump() const {
        const int d = get_max_depth();

        // If this tree is empty, tell someone
        if(d == 0) {
            cout << " <empty tree>\n";
            return;
        }

        // This tree is not empty, so get a list of node values...
        const auto rows_disp = get_row_display();
        // then format these into a text representation...
        auto formatted_rows = row_formatter(rows_disp);
        // then trim excess space characters from the left sides of the text...
        trim_rows_left(formatted_rows);
        // then dump the text to cout.
        for(const auto& row : formatted_rows) {
            std::cout << ' ' << row << '\n';
        }
    }
};


int main() {
    BinTree<int> bt;

    // Build OP's tree
    bt.insert(8,5,2,6,10,9,11);
    cout << "Tree from OP:\n\n";
    bt.Dump();
    cout << "\n\n";

    bt.clear();

    // Build a random tree 
    // This toy tree can't balance, so random
    // trees often look more like linked lists.
    // Just keep trying until a nice one shows up.
    std::random_device rd;
    std::mt19937 rng(rd());

    int MaxCount=20;
    int MaxDepth=5;
    const int Min=0, Max=1000;

    std::uniform_int_distribution<int> dist(Min,Max);

    while(MaxCount--) {
        bt.insert(dist(rng));
        if(bt.get_max_depth() >= MaxDepth) break;
    }

    cout << "Randomly generated tree:\n\n";
    bt.Dump();
}

输出示例:

Tree from OP:

       8
      / \
     /   \
    /     \
   5      10
  / \     / \
 2   6   9  11


Randomly generated tree:

                        703
                        / \
                       /   \
                      /     \
                     /       \
                    /         \
                   /           \
                  /             \
                 /               \
                /                 \
               /                   \
              /                     \
             /                       \
            /                         \
           /                           \
          /                             \
        137                             965
        / \                             /
       /   \                           /
      /     \                         /
     /       \                       /
    /         \                     /
   /           \                   /
  /             \                 /
 41             387             786
  \             / \             / \
   \           /   \           /   \
    \         /     \         /     \
    95      382     630     726     813
                                      \
                                      841