inorder

The problem I am tackle with is to find the first occurrence node in its inorder traversal in a BST. The code I have is given below def Inorder_search_recursive(node,key): if not node: return None InOrder_search_recursive(node.lChild) if node.value==key: return node InOrder_search_recursive(node.rChild) This code always return None, what's wrong with it. I think I've return node when I find a node with value k. Why cannot python pass this node???Thanks in advance When you call yourself recursively, like this: InOrder_search_recursive(node.lChild) That's just a normal function call, like any

I am writing a program to traverse a binary search tree.Here's my code: Main.java public class Main { public static void main(String[] args) { BinaryTree binaryTree = new BinaryTree(); binaryTree.add(50); binaryTree.add(40); binaryTree.add(39); binaryTree.add(42); binaryTree.add(41); binaryTree.add(43); binaryTree.add(55); binaryTree.add(65); binaryTree.add(60); binaryTree.inOrderTraversal(binaryTree.root); } } Node.java public class Node { int data; Node left; Node right; Node parent; public Node(int d) { data = d; left = null; right = null; } } BinaryTree.java public class BinaryTree { Node

I have written the following code for constructing a tree from its inorder and preorder traversals. It looks correct to me but the final tree it results in does not have the same inorder output as the one it was built from. Can anyone help me find the flaw in this function? public btree makeTree(int[] preorder, int[] inorder, int left,int right) { if(left > right) return null; if(preIndex >= preorder.length) return null; btree tree = new btree(preorder[preIndex]); preIndex++; int i=0; for(i=left; i<= right;i++) { if(inorder[i]==tree.value) break; } tree.left = makeTree(preorder, inorder,left,

I am working on translating an expression tree to a format that resembles infix notation; I am not evaluating the tree or executing its operations. The tree contains both logical and relational operations, and I would like to emit parentheses in an intelligent manner during the translation. To illustrate, consider the following contrived expression: a < x & (a < y | a == c) & a != d If I walk the expression tree produced by this expression in-order, then I will print out the following expression, which is incorrect. a < x & a < y | a == c & a != d // equivalent to (a < x & a < y) | (a == c & a