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public: void inord(TreeNode *root, vector <int> inorder){ if(!root) return; if(root->left) inord(root->left, inorder); inorder.push_back(root->val); if(root->right) inord(root->right, inorder); } vector<int> inorderTraversal(TreeNode* root) { // if(root==NULL) return {}; vector <int> inorder; inord(root, inorder); return inorder; }};

Question

public: void inord(TreeNode root, vector <int> inorder){ if(!root) return; if(root->left) inord(root->left, inorder); inorder.push_back(root->val); if(root->right) inord(root->right, inorder); } vector<int> inorderTraversal(TreeNode root) { // if(root==NULL) return {}; vector <int> inorder; inord(root, inorder); return inorder; }};

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Solution

This is a C++ function for an Inorder Traversal of a Binary Tree. Here's a step-by-step explanation:

  1. The function inord(TreeNode *root, vector <int> inorder) is a helper function that performs the actual Inorder Traversal. It takes a pointer to a TreeNode and a reference to a vector of integers as arguments.

  2. If the root is null, the function returns immediately. This is the base case for the recursion.

  3. If the root has a left child, the function calls itself recursively with the left child as the new root.

  4. After returning from the left subtree, it adds the value of

This problem has been solved

Similar Questions

What is wrong with the below code for the inorder traversal of in order threaded binary tree:

- Provide a detailed algorithm for the Inorder tree traversal method. explain the answer for 5 marks

Invert the Binary TreeGiven the root of a binary tree, invert the tree, and return its root. Print the Inorder traversal of the inverted tree.Input format:The input consists of one line, space-separated elements denoting the nodes of the Binary Tree.Output format:Print the inorder traversal of the inverted tree in a single line.

Complete below code of Preorder traversalPreorder(node* root){If(a-----------------){cout<<root->data;b--------------------------;preorder(root->right);}}                                                        a.NULL b.root->lefta.root==NULL b.Preorder(root->left);a.root!=NULL b.Preorder(root->left);a.root!=NULL b.root->left

Complete the function in the editor below. It received parameter: a pointer to the root of a binary tree. It must print the values in the tree's postorder traversal as a single line of space-separated values.

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