ADVANCED DATA STRUCTURES PROGRAM • LEVEL 14 — BALANCED BINARY SEARCH TREES

Insert into a Left-Leaning Red–Black Tree

Learn how to insert into a left-leaning red–black tree using a clear C program.

AdvancedSelf-balancing BSTLogarithmic operations

PROBLEM UNDERSTANDING

Input and expected output

Sample input
No input required
Sample output
10 15 20 25 30
Root = 20 Black = Yes

COMPLETE C PROGRAM

Complete C implementation

ads-red-black-left-leaning-insertion.c
Open in compiler
#include <stdio.h>
#include <stdlib.h>

struct Node { int key, red; struct Node *left, *right; };
int is_red(struct Node *node) { return node && node->red; }
struct Node *make_node(int key) { struct Node *node = malloc(sizeof *node); if (!node) exit(EXIT_FAILURE); node->key = key; node->red = 1; node->left = node->right = NULL; return node; }
struct Node *rotate_left_rb(struct Node *root) { struct Node *x = root->right; root->right = x->left; x->left = root; x->red = root->red; root->red = 1; return x; }
struct Node *rotate_right_rb(struct Node *root) { struct Node *x = root->left; root->left = x->right; x->right = root; x->red = root->red; root->red = 1; return x; }
void flip(struct Node *root) { root->red = !root->red; root->left->red = !root->left->red; root->right->red = !root->right->red; }
struct Node *insert_node(struct Node *root, int key) { if (!root) return make_node(key); if (key < root->key) root->left = insert_node(root->left, key); else if (key > root->key) root->right = insert_node(root->right, key); if (is_red(root->right) && !is_red(root->left)) root = rotate_left_rb(root); if (is_red(root->left) && is_red(root->left->left)) root = rotate_right_rb(root); if (is_red(root->left) && is_red(root->right)) flip(root); return root; }
void inorder_node(struct Node *root) { if (root) { inorder_node(root->left); printf("%d ", root->key); inorder_node(root->right); } }
void free_nodes(struct Node *root) { if (root) { free_nodes(root->left); free_nodes(root->right); free(root); } }

int main(void)
{
    int keys[] = {10, 20, 30, 15, 25}; struct Node *root = NULL;
    for (int i = 0; i < 5; i++) { root = insert_node(root, keys[i]); root->red = 0; }
    inorder_node(root); printf("\nRoot = %d Black = %s\n", root->key, root->red ? "No" : "Yes"); free_nodes(root); return 0;
}

GUIDED CODE TOUR • NOT LIVE EXECUTION

Study the program line by line

Use the real compiler button above to run and debug with different inputs.

CURRENT STEP

Select Start to walk through the important lines.

SELECTED LINE

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EXPECTED OUTPUT FOR THE SAMPLE

10 15 20 25 30
Root = 20 Black = Yes
0%Step 0 of 0

PROGRAM EXPLANATION

Algorithm and explanation

  1. Read the required input values.
  2. Use rotations and color flips to encode balanced 2–3 tree operations in a binary tree.
  3. Display the computed result.

Use rotations and color flips to encode balanced 2–3 tree operations in a binary tree.

EFFICIENCY

Time and space complexity

Time complexity

O(log n) per insertion

Auxiliary space

O(h)

DEBUGGING CHECKLIST

Common mistakes

Check this

Use the correct format specifier for every variable.

Check this

Initialize variables before using their values.

Check this

Check braces, semicolons and input order carefully.

Try it yourself

Practice: Run the program with the sample input, predict its output, and then test one boundary case of your own.