## Print binary tree in a pretty way using c++

pretty print binary tree java

c++ print binary tree level by level

java print tree recursion

python print binary tree

print binary tree javascript

print binary tree - leetcode

recursively print binary tree c++

I am a "bit" lost trying to print a binary tree like 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 in each level, but I couldn't figure out how to set the right number of spaces between the root and the second level (there are 3 lines under the root for 3 levels but I believe it is not this everytime,I thought it could be 3 times the height for greater trees).

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

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); }

Here is an example of code creating a text-based representation of a binary tree. This demonstration uses a minimally useful binary tree class (BinTree), with a small footprint, just to avoid bloating the example's size.

Its text-rendering member functions are more serious, using iteration rather than recursion, as found in other parts of the class.

This does its job in three steps, first a vector of rows of string values is put together.

Then this is used to format lines of text strings representing the tree.

Then the strings are cleaned up and dumped to cout.

As an added bonus, the demo includes a "random tree" feature, for hours of nonstop entertainment.

#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(); }

An example of the output:

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

**Print a binary tree in a pretty way,** Call Print with the root of your tree. From your root, count the number of your left children. From the total number of left children, proceed with printing the root with the indention of the number of left children. Pretty-printing a binary tree in C (and other imperative languages) (First-time poster and rather new in programming, so be patient, please!) I'm interested in both an efficient general algorithm for printing formatted binary trees (in a CLI environment) and a C implementation.

Even tough it is not exactly what you asked for, printing trees horizontally is way simpler. And especially in case of large trees, I think this is the better representation form.

└──8 ├──5 │ ├──2 │ └──6 └──10 ├──9 └──11

Following C++ code roots in this java implementation.

void printBT(const std::string& prefix, const BSTNode* node, bool isLeft) { if( node != nullptr ) { std::cout << prefix; std::cout << (isLeft ? "├──" : "└──" ); // print the value of the node std::cout << node->m_val << std::endl; // enter the next tree level - left and right branch printBT( prefix + (isLeft ? "│ " : " "), node->m_left, true); printBT( prefix + (isLeft ? "│ " : " "), node->m_right, false); } } void printBT(const BSTNode* node) { printBT("", node, false); } // pass the root node of your binary tree printBT(root);

**Print Binary Tree in 2-Dimensions,** Print Binary Tree in 2-Dimensions · Breadth First Search without using Queue · Check if a is subtree of another binary tree using preorder traversal : Iterative · Print path between So we do a reverse inorder traversal (right – root – left) and print tree nodes. C++. filter_none. edit close. play_arrow. link brightness_4 code Print Binary Tree Structure with its contents in C++. Given a binary tree, write an efficient algorithm to print binary tree structure in standard output. For example, binary tree to the left can be displayed as binary tree to the right programmatically. Below code serves as an excellent helper function to binary tree problems for printing the tree.

**Print Binary Tree Structure with its contents in C++,** Print Binary Tree Structure with its contents in C++. Given a binary tree, write an efficient algorithm to print binary tree structure in standard output. Given a binary tree, print it vertically. The following example illustrates vertical order traversal. The idea is to traverse the tree once and get the minimum and maximum horizontal distance with respect to root. For the tree shown above, minimum distance is -2 (for node with value 4) and maximum distance is 3 (For node with value 9).

**Print Binary Tree,** Print a binary tree in an m*n 2D string array following these rules: is not, you don't need to print anything for the none subtree but still need to leave the space The below routine prints tree in ascii for a given Tree representation which contains list of nodes, and node structure is this struct Tree { Tree * left, * right; int element; }; This pic illustrates what the below routine does on canvas.. ascii tree Here is the printing routine..

**Print a binary tree · GitHub,** Print a binary tree. print-binary-tree.c. void padding struct node *tree = NULL;. /* */ 90 degree rotation, I use this for printing the structure of a tree because it's fast and easy: I thought it would be nice to let you know. Check if a binary tree is subtree of another binary tree using preorder traversal : Iterative; Count nodes in the given tree whose weight is a fibonacci number; Print all pairs from two BSTs whose sum is greater than the given value; Split a BST into two balanced BSTs based on a value K; Print all Exponential Levels of a Binary Tree; Diamond Tree

**jdmcpeek/pretty-print-binary-tree: Simple python module ,** Simple python module that pretty-prints a binary tree to the command line. layer 1 root = Node('A') # layer 2 root.left = Node('B') root.right = Node('C') add a sub-tree ('X' nodes) using the createTree method root.left.right.right = Node. Refer below post for level order traversal based solution. The below post makes sure that nodes of a vertical line are printed in the same order as they appear in the tree. Print a Binary Tree in Vertical Order | Set 3 (Using Level Order Traversal)

**Print binary tree in a pretty way using c++,** Break Statement In C Programming Language In HINDIhttps://youtu.be/eq9NyUx_feI Break Duration: 7:43
Posted: Apr 20, 2020 The traditional way to do a breadth-first traversal of a tree is to use a queue, and to work iteratively on elements of the queue rather than recursively on elements of the tree. Not only is this a less complex data structure than an unordered map of ints to vectors of nodes, it also stores nodes from no more than two levels of the tree at a

##### Comments

- A binary tree may be printed in many different ways, see en.wikipedia.org/wiki/Tree_traversal. What is your desired output?
- Are you trying to print the binary tree on the terminal?
- I'd like to print it in the level-order,but I wanto print like a mentioned in the post, with the "/" ,jumping line and so on...
- The output must be a tree in the console, not a sequence of numbers
- Thank you Christopher. I think there is no easy way for this problem. I almost got a quick solution using queues but I couldn't figure out the intern distances.
- I was surprised at how involved it turned out to be, but I was interested in seeing it finished because I have wanted something like this on several occasions in the past.
- @ChristopherOicles. could you please mention the compilation dependencies? using -std=c++11 flag throwing a lot of warnings+errors in my windows system.
- @Debashish You probably have an old version of gcc which doesn't fully support c++11, it probably lacks the
`random`

library and may not support`using`

for typedefs.