Bottom-left to upward-right Traversal in a Binary Tree
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 <bits/stdc++.h>#include <iostream>using namespace std;// Node Structurestypedef struct Node { int data; Node* left; Node* right;} Node;// Function to add the new Node in// the Binary TreeNode* 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 ordervector<int>leftBottomTopRightTraversal(Node* root){ // Stores the data of the node vector<int> rr; // Stores every element in each layer stack<int> r; // Stores the roots in the layers queue<Node*> 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 stringNode* buildBinaryTree(char* t){ Node* root = NULL; // Using queue to build tree queue<Node**> 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 Codeint 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<int> result = leftBottomTopRightTraversal(root); // Print the final traversal for (int i = 0; i < result.size(); ++i) { cout << result[i] << " "; } return 0;} |
2 7 2 5 6 5 11 4 9
Time Complexity: O(N)
Auxiliary Space: O(N)
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.
In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.
