Printing a binary tree nicely using C ++

Question

I lost the “bit” while trying to print a binary tree, as shown below in C ++:

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

I know how to get the height of the tree and the number of nodes at each level, but I could not figure out how to set the correct number of spaces between the root and the second level (there are 3 lines under the root for 3 levels, but I believe that this is not so every time I thought it could be 3 times the height for large trees).

I would like to have some help to print these spaces in lines and the number of lines between lines. Thanks.

I am coding in C ++

 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

Here is an example of code that creates a text representation of a binary tree. This demo uses a minimally usable binary tree class (BinTree) with a small footprint to avoid inflating the size of the example.

Its text display functions are more serious using iteration rather than recursion, as shown in other parts of the class.

This does its job in three stages: first, the string vector of string values ​​is carried over.

This is then used to format the lines of text lines representing the tree.

Then the lines are cleared and flushed to cout.

As an added bonus, the demo includes a "random tree" feature for several hours non-stop.

 #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 location, // its boolean "present" field is set to false. // The struct also contains a string representation of // its Node 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 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 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(); } 

Output Example:

 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 

Answer 2

• 1st function - prints a level by level (root lv → lv)
• 2nd function - distance from the beginning of a new line
• 3rd function - prints nodes and calculates the distance between two prints;

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 void Tree::TREEPRINT() { int i = 0; while (i <= treeHeight(getroot())){ printlv(i); i++; cout << endl; } } void Tree::printlv(int n){ Node* temp = getroot(); int val = pow(2, treeHeight(root) -n+2); cout << setw(val) << ""; prinlv(temp, n, val); } void Tree::dispLV(Node*p, int lv, int d) { int disp = 2 * d; if (lv == 0){ if (p == NULL){ cout << " x "; cout << setw(disp -3) << ""; return; } else{ int result = ((p->key <= 1) ? 1 : log10(p->key) + 1); cout << " " << p->key << " "; cout << setw(disp - result-2) << ""; } } else { if (p == NULL&& lv >= 1){ dispLV(NULL, lv - 1, d); dispLV(NULL, lv - 1, d); } else{ dispLV(p->left, lv - 1, d); dispLV(p->right, lv - 1, d); } } } 

Input:

 50-28-19-30-29-17-42-200-160-170-180-240-44-26-27 

Exit: https://i.stack.imgur.com/TtPXY.png