{"id":563,"date":"2024-10-19T13:53:29","date_gmt":"2024-10-19T08:23:29","guid":{"rendered":"https:\/\/codexplained.in\/?p=563"},"modified":"2025-11-24T15:32:30","modified_gmt":"2025-11-24T10:02:30","slug":"inorder-traversal-of-binary-tree","status":"publish","type":"post","link":"https:\/\/codexplained.in\/?p=563","title":{"rendered":"Inorder Traversal of Binary Tree"},"content":{"rendered":"
\n#include <stdio.h>\n#include <stdlib.h>\n\n\/\/ Define the structure of a binary tree node\nstruct Node {\n    int data;\n    struct Node* left;\n    struct Node* right;\n};\n\n\/\/ Function to create a new node\nstruct Node* createNode(int value) {\n    struct Node* newNode = (struct Node*)malloc(sizeof(struct Node));\n    newNode->data = value;\n    newNode->left = NULL;\n    newNode->right = NULL;\n    return newNode;\n}\n\n\/\/ Function for inorder traversal\nvoid inorderTraversal(struct Node* root) {\n    if (root == NULL) {\n        return;\n    }\n\n    \/\/ Recursively traverse the left subtree\n    inorderTraversal(root->left);\n\n    \/\/ Visit the root node\n    printf("%d ", root->data);\n\n    \/\/ Recursively traverse the right subtree\n    inorderTraversal(root->right);\n}\n\nint main() {\n    \/\/ Creating nodes of the binary tree\n    struct Node* root = createNode(1);\n    root->left = createNode(2);\n    root->right = createNode(3);\n    root->left->left = createNode(4);\n    root->left->right = createNode(5);\n\n    printf("Inorder traversal of the binary tree is: ");\n    inorderTraversal(root);\n    printf("\\n");\n\n    return 0;\n}\n\n<\/pre><\/div>\n\n\n
Explanation of the Code:<\/h3>\n\n\n\n
    \n
  1. Structure Definition<\/strong>:\n
      \n
    • We define a structure Node<\/code> to represent a node of the binary tree.<\/li>\n\n\n\n
    • Each node contains three elements:\n
        \n
      • data<\/code>: Holds the value of the node.<\/li>\n\n\n\n
      • left<\/code>: Pointer to the left child.<\/li>\n\n\n\n
      • right<\/code>: Pointer to the right child.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n\n\n\n
      • createNode() Function<\/strong>:\n
          \n
        • This function allocates memory for a new node, sets the data<\/code> field, and initializes the left and right pointers to NULL<\/code>.<\/li>\n\n\n\n
        • It returns a pointer to the newly created node.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
        • inorderTraversal() Function<\/strong>:\n
            \n
          • This function takes the root node of a binary tree as input.<\/li>\n\n\n\n
          • It uses recursion to traverse the tree in the following order:\n
              \n
            • First, it calls inorderTraversal()<\/code> for the left subtree.<\/li>\n\n\n\n
            • Then, it prints the data<\/code> of the current node.<\/li>\n\n\n\n
            • Finally, it calls inorderTraversal()<\/code> for the right subtree.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
            • If the root<\/code> is NULL<\/code> (base condition), it simply returns, ending that branch of recursion.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
            • main() Function<\/strong>:\n
                \n
              • This function serves as the entry point of the program.<\/li>\n\n\n\n
              • We create a sample binary tree with 5 nodes.<\/li>\n\n\n\n
              • The tree looks like this:<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n
                     1\n    \/ \\\n   2   3\n  \/ \\\n 4   5\n<\/code><\/pre>\n\n\n\n
                Output:<\/h3>\n\n\n\n
                When the program is run, the output will be:<\/p>\n\n\n\n
                Inorder traversal of the binary tree is: 4 2 5 1 3\n<\/code><\/pre>\n\n\n\n
                Explanation of the Output:<\/h3>\n\n\n\n
                For the given binary tree:<\/p>\n\n\n\n
                  \n
                1. Start at the root (1)<\/strong>:\n
                    \n
                  • Traverse the left subtree of 1<\/code> (which is 2<\/code>).<\/li>\n<\/ul>\n<\/li>\n\n\n\n
                  • At node 2<\/strong>:\n
                      \n
                    • Traverse the left subtree of 2<\/code> (which is 4<\/code>).<\/li>\n\n\n\n
                    • 4<\/code> is a leaf node (no children), so we print 4<\/code>.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
                    • Back to node 2<\/strong>:\n
                        \n
                      • After visiting 4<\/code>, we print 2<\/code>.<\/li>\n\n\n\n
                      • Then, we traverse the right subtree of 2<\/code> (which is 5<\/code>).<\/li>\n<\/ul>\n<\/li>\n\n\n\n
                      • At node 5<\/strong>:\n
                          \n
                        • 5<\/code> is a leaf node, so we print 5<\/code>.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
                        • Back to the root (1)<\/strong>:\n
                            \n
                          • After finishing the left subtree, we print 1<\/code>.<\/li>\n\n\n\n
                          • Then, traverse the right subtree of 1<\/code> (which is 3<\/code>).<\/li>\n<\/ul>\n<\/li>\n\n\n\n
                          • At node 3<\/strong>:\n
                              \n
                            • 3<\/code> is a leaf node, so we print 3<\/code>.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n
                              Thus, the output of the inorder traversal is 4 2 5 1 3<\/code>.<\/p>\n