[ad_1] You are passing prefix_exp_str[1:] to both the recursive calls. Instead, the second recursive call needs to receive whatever is left of the input string after the first recursive call is finished. [ad_2] solved Creating an expression tree for prefix expression
[ad_1] void reverseLevelOrder(struct node* root) { int h = height(root); int i; for (i=h; i>=1; i–) //THE ONLY LINE DIFFERENT FROM NORMAL LEVEL ORDER printGivenLevel(root, i); } /* Print nodes at a given level */ void printGivenLevel(struct node* root, int level) { if (root == NULL) return; if (level == 1) printf(“%d “, root->data); else … Read more
[ad_1] int countUniVals(node* head, bool* unival) { if (!node) { *unival = true; return 0; } bool uniL,uniR; int sum = countUniVals(head->l, &uniL) + countUniVals(head->r, &uniR); if (uniL && uniR && (!head->l || head->l->val == head->val) && (!head->r || head->r->val == head->val)) { sum++; *unival = true; } return sum; } 0 [ad_2] solved Counting … Read more
Introduction [ad_1] A unival subtree is a tree where all the nodes have the same value. Counting the number of unival subtrees in a binary tree is a common problem in computer science. This problem can be solved using a recursive approach, which involves traversing the tree and checking each node to see if it … Read more
[ad_1] Why isn’t std::set just called std::binary_tree? Because Tree doesn’t describe how the interface is used. Set does. std::set does not provide sufficient operations to be used as a general purpose search tree. It only provides an interface to a particular application of a search tree: The representation of a set. Technically the standard doesn’t … Read more
[ad_1] It is a nuisance when we have to create an MCVE from fragments of a program. However, it’s doable — it’s a pain, but doable. Here’s my variation on your code. Of the 115 lines shown, 35 are what you wrote in the question — roughly. So, I had to create twice as much … Read more
[ad_1] The pre order goes like this: Visit the current node If exists left node, visit pre order of left node. If exist right node, visit pre order of right node. I’m sure that with these few steps you will be able to implement your pre order method. A pseudocode of this: function preOrder(Node node) … Read more