Print the first shortest root to leaf path in a Binary Tree in C++ Programming.

C++Server Side ProgrammingProgramming

Given the binary tree the program must find out the shortest path from the root to leaf amongst many given paths.

Since we traverse the tree from left to right, so if there are multiple shortest paths from the root to leaf than the program will print the first traversed the shortest path on the left side of a tree.

We can use a queue that will traverse each level using Level order traversal and the path with the least number of levels will be printed as it will be the shortest path from the root to leaf

In the above tree, multiple paths from the root to leaf are

10 -> 3 (this one is the shortest path amongst all)
10 -> 211 -> 100
10 -> 211 -> 146

Example

Input  : 10 3 211 100 146
Output : 10 3

Algorithm

Step 1 -> create a structure of a node as
   struct node
      struct node *left, *right
      int data
   End
Step 2 -> function to create a node
   node* newnode(int data)
   node *temp = new node
   temp->data = data
   temp->left = temp->right= NULL
   return temp
Step 3 -> create function for calculating path
   void path(int data, unordered_map <int,int> prnt)
      IF prnt[data] = data
         Return
      End
      path(prnt[data], prnt)
      print prnt[data]
step 4 -> function for finding out the left path
   void left(Node* root)
   create STL queue<Node*> que
   que.push(root)
   int leaf = -1
   Node* temp = NULL
   Create STL unordered_map<int, int> prnt
   prnt[root->data] = root->data
   Loop While !que.empty()
      temp = que.front()
      que.pop()
      IF !temp->left && !temp->right
         leaf = temp->data
         break
      End
      Else
         IF temp->left
            que.push(temp->left)
            prnt[temp->left->data] = temp->data
         End
         IF temp->right
            que.push(temp->right)
            prnt[temp->right->data] = temp->data
         End
      End
   End
   path(leaf, prnt)
   print leaf
Step 5 -> In main()
   Create tree using Node* root = newnode(90)
   root->left = newnode(21)
   call left(root)
stop

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
// structure of a node
struct Node {
   struct Node *left,*right;
   int data;
};
//function to create a new node
Node* newnode(int data){
   Node* temp = new Node;
   temp->data = data;
   temp->left = NULL;
   temp->right = NULL;
   return temp;
}
//function to set a path
void path(int data, unordered_map <int,int> prnt) {
   if (prnt[data] == data)
      return;
   path(prnt[data], prnt);
   cout << prnt[data] << " ";
}
//function for a leaf path
void left(Node* root) {
   queue<Node*> que;
   que.push(root);
   int leaf = -1;
   Node* temp = NULL;
   unordered_map<int, int> prnt;
   prnt[root->data] = root->data;
   while (!que.empty()){
      temp = que.front();
      que.pop();
      if (!temp->left && !temp->right{
         leaf = temp->data;
         break;
      } else {
         if (temp->left){
            que.push(temp->left);
            prnt[temp->left->data] = temp->data;
         }
         if (temp->right){
            que.push(temp->right);
            prnt[temp->right->data] = temp->data;
         }
      }
   }
   path(leaf, prnt);
   cout << leaf << " ";
}
int main(){
   Node* root = newnode(90);
   root->left = newnode(21);
   root->right = newnode(32);
   root->left->left = newnode(45);
   root->right->left = newnode(52);
   root->right->right = newnode(27);
   root->left->left->left = newnode(109);
   root->left->left->right = newnode(101);
   root->right->right->left = newnode(78);
   left(root);
   return 0;
}

Output

if we run the above program then it will generate the following output

90 32 52
raja
Published on 04-Sep-2019 10:27:54
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