Program to find the left side view of a binary tree in C++

Suppose we have a binary tree, if we see the tree from left side, then we can see some elements of it. we have to display those elements. So if the tree is like −

The output will be [1,2,5]

To solve this, we will follow these steps −

• Define an array ret

• Define a function dfs(), this will take node, c initialize it with 1,

• if node is null, then −

  • return

• if c > lvl, then −

  • lvl := c

  • insert value of node into ret

• dfs(left of node, c + 1)

• dfs(right of node, c + 1)

• From the main method, do the following −

• lvl := -1

• dfs(root, 0)

• return ret

  
  

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<int> v){
   cout << "[";
   for(int i = 0; i<v.size(); i++){
      cout << v[i] << ", ";
   }
   cout << "]"<<endl;
}
class TreeNode{
   public:
   int val;
   TreeNode *left, *right;
   TreeNode(int data){
      val = data;
      left = right = NULL;
   }
};
class Solution {
   public:
   vector <int> ret;
   int lvl;
   void dfs(TreeNode* node, int c = 1){
      if(!node)
         return;
      if(c > lvl){
         lvl = c;
         ret.push_back(node->val);
      }
      dfs(node->left, c + 1);
      dfs(node->right, c + 1);
   }
   vector<int> solve(TreeNode* root) {
      lvl = -1;
      dfs(root, 0);
      return ret;
   }
};
main(){
   TreeNode *root = new TreeNode(1);
   root->left = new TreeNode(2);
   root->right = new TreeNode(3);
   root->left->right = new TreeNode(5);
   root->right->right = new TreeNode(4);
   Solution ob;
   print_vector(ob.solve(root));
}

Input

TreeNode *root = new TreeNode(1);
root->left = new TreeNode(2);
root->right = new TreeNode(3);
root->left->right = new TreeNode(5);
root->right->right = new TreeNode(4);

Output

[1,2,5]