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Anti-Clockwise spiral traversal of a binary tree in C++

In this article, you will learn about the anti-clockwise spiral traversal of a binary tree in C++ with its algorithm and example.


Within the field of tree traversals, the anti-clockwise spiral traversal offers a visually appealing and thought-provoking look at the architecture of a binary tree. It winds its way around the tree in a mesmerizing dance, defying convention by going against the usual left-to-right or right-to-left directions and visiting nodes in a distinctive spiralling pattern.


Initialize variables:

  • i: Starts at 1, which represents the level to begin from (bottom level for anti-clockwise spiral).
  • j: Starts at the height of the tree, which represents the level to end at (root level).
  • flag: Boolean variable, which is initially set to false to indicate right-to-left traversal.

Loop until all levels are covered:

  • Check the flag value:
  • flag is true (left-to-right):
  • Traverse levels from i (current bottom level) to j (current top level), printing node data at each level.
  • flag is false (right-to-left):
  • Traverse levels from j (current top level) down to i (current bottom level), printing node data at each level.
  • Change the flag value to its opposite for the next level.
  • Update i and j according to the flag value:
  • flag is true (left-to-right):Increment i to move down to the next level.
  • Decrement j to exclude the top level for left-to-right traversal.
  • flag is false (right-to-left):
  • Increment i to move down to the next level for right-to-left traversal.

Data Structures:

No stacks or queues are explicitly required for this algorithm.

Auxiliary variables:

i, j: These variable track the current and ending levels for traversal.

flag: It indicates the direction of traversal (left-to-right or right-to-left).


The provided C++ code implements the algorithm with these improvements:

  • Error handling: It checks for empty trees and invalid level values.
  • printNode function: It replaces with your specific logic to print node data at a given level (might involve level order traversal for larger trees).


Let us take an example to illustrate the anti-clockwise spiral traversal of a binary tree in C++.




using namespace std;

struct TreeNode {

    int value;

    TreeNode* left;

    TreeNode* right;

    TreeNode(int value) : value(value), left(nullptr), right(nullptr) {}


int calculateHeight(TreeNode* root);

void printSpiralLevel(TreeNode* root, int level, bool leftToRight);

void spiralTraversal(TreeNode* root) {

    if (!root) {



    int treeHeight = calculateHeight(root);

    bool leftToRight = false;

    for (int currentLevel = 1; currentLevel <= treeHeight; currentLevel++) {

        printSpiralLevel(root, currentLevel, leftToRight);

        leftToRight = !leftToRight;



int calculateHeight(TreeNode* root) {

    if (!root) {

        return 0;


    int leftHeight = calculateHeight(root->left);

    int rightHeight = calculateHeight(root->right);

    return max(leftHeight, rightHeight) + 1;


void printSpiralLevel(TreeNode* root, int level, bool leftToRight) {

    if (!root) {



    if (level == 1) {

        cout << root->value << " ";

    } else if (level > 1) {

        if (leftToRight) {

            printSpiralLevel(root->left, level - 1, leftToRight);

            printSpiralLevel(root->right, level - 1, leftToRight);

        } else {

            printSpiralLevel(root->right, level - 1, leftToRight);

            printSpiralLevel(root->left, level - 1, leftToRight);




TreeNode* constructTree() {

    int rootValue;

    cout << "Enter value for root node: ";

    cin >> rootValue;

    TreeNode* root = new TreeNode(rootValue);

    queue nodesQueue;


    while (!nodesQueue.empty()) {

        TreeNode* currentNode = nodesQueue.front();


        int leftValue, rightValue;

        cout << "Enter left child value of " << currentNode->value << " (or -1 if none): ";

        cin >> leftValue;

        if (leftValue != -1) {

            currentNode->left = new TreeNode(leftValue);



        cout << "Enter right child value of " << currentNode->value << " (or -1 if none): ";

        cin >> rightValue;

        if (rightValue != -1) {

            currentNode->right = new TreeNode(rightValue);




    return root;


int main() {

    TreeNode* root = constructTree();


    return 0;



Explanation of the code:

  • In this example, the antiClockwiseSpiralTraversal function takes the root node of the binary tree as input.
  • After that, the height function calculates the height of the tree efficiently.
  • Next, the printNode function is a placeholder that replace it with your actual printing logic, depending on your requirements.
  • The main loop iterates until all levels (i <= j) are covered.
  • Inside the loop, the flag value determines the direction of traversal.
  • Levels are traversed, and nodes are printed according to the direction and current level.
  • The flag and level indicators are updated for the next iteration.

Time Complexity Analysis:

Constructing the Tree:

  • The amount of nodes in the tree determines how time-consuming it is to build.
  • The user enters the left and right child values of each node, yielding a linear time complexity of O(N), where N is the total number of nodes.

Calculating Tree Height:

  • The calculateHeight function iteratively goes through the whole tree to determine the height of a tree.
  • The temporal complexity is O(N), where N is the number of nodes because each node is visited once.

Spiral Traversal:

  • The spiral traversal function goes through the tree's levels one by one until it reaches its peak.
  • The printSpiralLevel function is executed for each level, traversing the level's nodes.
  • In the worst-case scenario, when all nodes are visited once, the time complexity is O(N), where N is the total number of nodes.
  • Overall, the time complexity of the code is O(N), where N is the number of nodes in the tree.

Space Complexity Analysis:

Tree Construction:

  • The queue to build trees and the tree nodes must be stored somewhere.
  • This section has an O(N) space complexity, where N is the number of nodes in the tree.

Calculating Tree Height:

  • The tree height affects the space complexity of the recursive calls to calculateHeight.
  • The space complexity is O(H), where H is the tree's height in the worst scenario when the tree is skewed.

Spiral Traversal:

  • The tree's height also affects the space complexity during the spiral traversal.
  • The space complexity is O(H) in the worst scenario, where the tree is skewed.

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