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Boundary Traversal of binary tree.cpp
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Boundary Traversal of binary tree.cpp
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/*
Solution by Rahul Surana
***********************************************************
Given a Binary Tree, find its Boundary Traversal. The traversal should be in the following order:
Left boundary nodes: defined as the path from the root to the left-most node
ie- the leaf node you could reach when you always travel preferring the left subtree over the right subtree.
Leaf nodes: All the leaf nodes except for the ones that are part of left or right boundary.
Reverse right boundary nodes: defined as the path from the right-most node to the root.
The right-most node is the leaf node you could reach when you always travel preferring the right subtree over the left subtree.
Exclude the root from this as it was already included in the traversal of left boundary nodes.
Note: If the root doesn't have a left subtree or right subtree, then the root itself is the left or right boundary.
***********************************************************
*/
// { Driver Code Starts
#include <bits/stdc++.h>
using namespace std;
#define MAX_HEIGHT 100000
// Tree Node
struct Node
{
int data;
Node* left;
Node* right;
};
// Utility function to create a new Tree Node
Node* newNode(int val)
{
Node* temp = new Node;
temp->data = val;
temp->left = NULL;
temp->right = NULL;
return temp;
}
// Function to Build Tree
Node* buildTree(string str)
{
// Corner Case
if(str.length() == 0 || str[0] == 'N')
return NULL;
// Creating vector of strings from input
// string after spliting by space
vector<string> ip;
istringstream iss(str);
for(string str; iss >> str; )
ip.push_back(str);
// Create the root of the tree
Node* root = newNode(stoi(ip[0]));
// Push the root to the queue
queue<Node*> queue;
queue.push(root);
// Starting from the second element
int i = 1;
while(!queue.empty() && i < ip.size()) {
// Get and remove the front of the queue
Node* currNode = queue.front();
queue.pop();
// Get the current node's value from the string
string currVal = ip[i];
// If the left child is not null
if(currVal != "N") {
// Create the left child for the current node
currNode->left = newNode(stoi(currVal));
// Push it to the queue
queue.push(currNode->left);
}
// For the right child
i++;
if(i >= ip.size())
break;
currVal = ip[i];
// If the right child is not null
if(currVal != "N") {
// Create the right child for the current node
currNode->right = newNode(stoi(currVal));
// Push it to the queue
queue.push(currNode->right);
}
i++;
}
return root;
}
// } Driver Code Ends
/* A binary tree Node
struct Node
{
int data;
Node* left, * right;
}; */
class Solution {
public:
void left(Node* root, vector<int>& ans){
if(root == NULL) return;
if(root->left == NULL && root->right == NULL) return;
ans.push_back(root->data);
if(root->left) left(root->left,ans);
else if(root->right) left(root->right,ans);
}
void right(Node* root, vector<int>& ans){
if(root == NULL) return;
if(root->left == NULL && root->right == NULL) return;
if(root->right) right(root->right,ans);
else if(root->left) right(root->left,ans);
ans.push_back(root->data);
}
void leaf(Node* root, vector<int>& ans){
if(root == NULL) return;
if(root->left == NULL && root->right == NULL) {
ans.push_back(root->data);
return;
}
leaf(root->left,ans);
leaf(root->right,ans);
}
vector<int> printBoundary(Node *root)
{
vector<int> ans;
ans.push_back(root->data);
left(root->left,ans);
leaf(root,ans);
right(root->right,ans);
return ans;
}
};
// { Driver Code Starts.
/* Driver program to test size function*/
int main() {
int t;
string tc;
getline(cin, tc);
t=stoi(tc);
while(t--)
{
string s ,ch;
getline(cin, s);
Node* root = buildTree(s);
Solution ob;
vector <int> res = ob.printBoundary(root);
for (int i : res) cout << i << " ";
cout << endl;
}
return 0;
} // } Driver Code Ends