Bottom-left to upward-right Traversal in a Binary Tree

• Difficulty Level : Expert
• Last Updated : 15 Sep, 2021

Given a Binary Tree, the task is to print the Bottom-left to Upward-right Traversal of the given Binary Tree i.e., the level order traversal having level as Bottom-left to Upward-right node.

Examples:

Input: Below is the given Tree: Output: 2 7 2 5 6 5 11 4 9
Explanation:

Level 1: 2 7 2 (going upwards from bottom left to right to root)
Level 2: 5 6 5 (right from each node in layer 1/or bottom left to upwards right in this layer)
Level 3: 11 4 9 (right from each node in layer 2/or bottom left to upwards right in this layer)

Input: 1 2 3 4 5 6 7
Output: 4 2 1 5 6 3 2
Explanation
Layer 1: 4 2 1 (going upwards from bottom left to right to root)
Layer 2: 5 6 3 (right from each node in layer 1/or bottom left to upwards right in this layer)
Layer 3: 2 (right from each node in layer 2/or bottom left to upwards right in this layer)

Approach: The idea is to use the Breadth-First Search technique. Follow the steps needed to solve this problem:

• Initialize a layer in a binary tree. It is a list of nodes starting from the bottom-left most node next to the previous layer and ends with the upper-right most node next to the previous layer.
• Create a stack to stores all nodes in every layer.
• Initialize a queue to maintain “roots” in each layer, a root in a layer is a node from which one may go downwards using left children only.
• Push the root node of the first layer (the tree root) in the queue.
• Define an indicator (say lyr_root) a node expected at the end of a layer which is the current layer head, a layer head is the first node in a layer.
• Traverse until the queue is nonempty and do the following:
• Get a layer root from the front of the queue
• If this layer root is the layer head of a new layer, then, pop every element in the stack i.e., of the previous layer element, and print it.
• Traverse the layer from the upper-right to the bottom-left and for each element, if it has a right child, then check if the traversed node is the layer head or not. If found to be true then, change the expected indicator to indicate to the next layer head.
• Push the right child to the root in the queue.
• Push the traversed node in the stack.
• After traversing all the layers, the final layer may still be in the stack, so we need to pop every element from it and print it.

Below is the implementation of the above approach:

C++

 // C++ program for the above approach #include #include using namespace std;   // Node Structures typedef struct Node {     int data;     Node* left;     Node* right; } Node;   // Function to add the new Node in // the Binary Tree Node* newNode(int data) {     Node* n;       // Create a new Node     n = new Node();     n->data = data;     n->right = NULL;     n->left = NULL;     return n; }   // Function to traverse the tree in the // order of bottom left to the upward // right order vector leftBottomTopRightTraversal(Node* root) {     // Stores the data of the node     vector rr;       // Stores every element in each layer     stack r;       // Stores the roots in the layers     queue roots;       // Push the layer head of the     // first layer     roots.push(root);       // Define the first layer head     // as the tree root     Node* lyr_root = root;       // Traverse all layers     while (!roots.empty()) {           // get current layer root         Node* n = roots.front();           // Pop element from roots         roots.pop();         if (lyr_root == n) {               // Layer root was also             // the layer head             while (!r.empty()) {                   rr.push_back(r.top());                   // Pop every element                 // from the stack                 r.pop();             }         }           while (n) {               if (n->right) {                   // Current traversed node                 // has right child then                 // this root is next layer                 if (n == lyr_root) {                     lyr_root = n->right;                 }                   // Push the right child                 // to layer roots queue                 roots.push(n->right);             }               // Push node to the             // layer stack             r.push(n->data);             n = n->left;         }     }       // Insert all remaining elements     // for the traversal     while (!r.empty()) {           // After all of the layer         // roots traversed check the         // final layer in stack         rr.push_back(r.top());         r.pop();     }       // Return the traversal of nodes     return rr; }   // Function that builds the binary tree // from the given string Node* buildBinaryTree(char* t) {     Node* root = NULL;       // Using queue to build tree     queue q;     int data = 0;       // Stores the status of last     // node to be ignored or not     bool ignore_last = false;     while (*t != '\0') {         int d = *t - '0';           // If the current character         // is a digits then form the         // number of it         if (d >= 0 && d <= 9) {             data *= 10;             data += d;             ignore_last = false;         }           // If the current character         // is N then it is the         // NULL node         else if (*t == 'N') {             data = 0;             q.pop();             ignore_last = true;         }           // If space occured then         // add the number formed         else if (*t == ' ') {               // If last is ignored             if (!ignore_last) {                   // If root node is not NULL                 if (root) {                       Node** p = q.front();                     q.pop();                       if (p != NULL) {                         *p = newNode(data);                         q.push(&((*p)->left));                         q.push(&((*p)->right));                     }                 }                   // Else create a new                 // root node                 else {                     root = newNode(data);                     q.push(&(root->left));                     q.push(&(root->right));                 }                 data = 0;             }         }           // Increment t         t++;     }       // Return the root node of the tree     return root; }   // Driver Code int main() {     // Given order of nodes     char T[] = "2 7 5 2 6 N 9 N N 5 11 4 N";       // Builds the Binary Tree     Node* root = buildBinaryTree(T);       // Function Call     vector result         = leftBottomTopRightTraversal(root);       // Print the final traversal     for (int i = 0; i < result.size(); ++i) {         cout << result[i] << " ";     }       return 0; }

Output:

2 7 2 5 6 5 11 4 9

Time Complexity: O(N)
Auxiliary Space: O(N)

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