Dr Antonio Gulli - A collection of Tree Programming Interview Questions Solved in C++
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A collection of Tree Programming Interview Questions Solved in C++ Antonio Gulli Copyright 2015 Antonio Gulli All rights reserved. ISBN: ISBN-13: Tree is the fifth of a series of 25 Chapters devoted to algorithms, problem solving and C++ programming. DEDICATION To my wife Francesca Love always finds a path between forgiveness and karma ACKNOWLEDGMENTS Table of Contents
- First the current node is visited.
- Then the left children are pushed until a leaf is reached
- Then if the stack is empty, we leave the loop. Implementing an in-order visit for a Binary TreeSolutionA non-recursive in-order visit can be implemented by using a stack.Implementing an in-order visit for a Binary TreeSolutionA non-recursive in-order visit can be implemented by using a stack.
In a loop:
- First the left children are pushed until a leaf is reached
- Then if the stack is empty, we leave the loop. Otherwise the top of the stack pops out and the current node is visited.
- Finally the visit continues towards the right children
Codetemplate < typename tVal > void nonRecursiveInOrder( binaryTree < tVal > * root ) { std:: stack < binaryTree < tVal > *> s; while ( true ) { //left while ( root ) { s.push( root ); root = root ->left; } if (s.empty()) break ; // this one root = s.top(); s.pop(); std::cout << " v=" << root ->v__; //right root = root ->right; } }ComplexityTime complexity is
and space complexity is Implementing a post-order visit for a Binary TreeSolutionDuring post-ordering every single node is visited twice: the first time when moving towards the left children and then again, when moving towards the right children. In order to differentiate the two cases, we can compare if the current element and the right child of the element at the top of the stack are the same.Codetemplate < typename tVal > void nonRecursivePostOrder( binaryTree < tVal > * root ) { std:: stack < binaryTree < tVal > *> s; while ( true ) { if ( root ) { s.push( root ); root = root ->left; } else { if (s.empty()) break ; else if (!s.top()->right) { root = s.top(); s.pop(); std::cout << " v=" << root ->v__; if ( root == s.top()->right) { std::cout << " v=" << s.top()->v__; s.pop(); } } if (!s.empty()) root = s.top()->right; else root = NULL ; } // else } // while }ComplexityTime complexity is and space complexity is Implementing a level order visit for a Binary TreeSolutionLevel order visits can be implemented for all nodes at one level before going to the next level. The idea is very simple.Counting the number of leaves in a treeSolutionA solution can be provided by modifying the level order visit where we increment a counter every time we reach a leaf node.Codetemplate < typename tVal > unsigned NumLeavesNonRecursiveLevelOrder( binaryTree < tVal > * root ) { std:: queue < binaryTree < tVal > *> q; binaryTree < tVal > * tmp; unsigned count = 0; if (! root ) return count; q.push( root ); while (!q.empty()) { tmp = q.front(); q.pop(); if (!tmp->left && !tmp->right) count++; // another leaf else { if (tmp->left) q.push(tmp->left); if (tmp->right) q.push(tmp->right); } } return count; }ComplexityTime complexity is and space complexity is . Checking if two binary trees are structurally identicalSolutionA solution can be provided by recursion.Checking if two binary trees are structurally identicalSolutionA solution can be provided by recursion.We consider as base cases when both nodes are null, or if only one of them is null. Otherwise the algorithm checks if both nodes contain the same value and if both left children and right children are recursively structurally identical.
Codetemplate < typename tVal > unsigned structIdentical( binaryTree < tVal > * r1 , binaryTree < tVal > * r2 ) { if (! r1 && ! r2 ) return true ; if (! r1 || ! r2 ) return false ; return ( r1 ->v__ == r2 ->v__ && structIdentical( r1 ->left, r2 ->left) && structIdentical( r1 ->right, r2 ->right));}ComplexityTime complexity is and space complexity is . Printing all the paths in a binary treeSolutionA solution can be provided by recursion. The idea is to keep a vector
passing during recursion, while every node is pushed_back() to 
. Codetemplate < typename tVal > void printAllPaths( binaryTree < tVal > * root , std:: vector < tVal > path ) { if (! root ) return ; path .push_back( root ->v__); if (! root ->left && ! root ->right) { std:: vector < tVal >::iterator it = path.begin(); for (; it != path .end(); it++) std::cout << " v=" << *it; std::cout << std::endl; } else { printAllPaths( root ->left, path ); printAllPaths( root ->right, path ); } }ComplexityTime complexity is and space complexity is . Verifying if a path sum is equal to an integerSolutionThe idea is to subtract the current value from the given sum, while visiting recursively the tree.Codetemplate < typename tVal > bool hasSum( binaryTree < tVal > * root , int sum ) { if (! root ) return ( sum == 0); else { int remainingSum = sum - root ->v__; if (( root ->left && root ->right) || (! root ->left && ! root ->right)) return (hasSum( root ->left, remainingSum) || hasSum( root ->right, remainingSum)); else if ( root ->left) return hasSum( root ->left, remainingSum); else return hasSum( root ->right, remainingSum); } }ComplexityTime complexity is and space complexity is . Find the maximum path sum between two leaves of a binary treeSolutionThe maximum path sum between two leaves can either touch the root of the three or can be located on a subtree only if it belongs to the tree itself.
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