Solution #199 - Daniel/Edited - 17.03.2025

Trees (Binary Trees)/Views of Trees/Left and Right View/Explanation.md

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+Left View of tree.
+
+Brute Force Approach:
+
+1. Perform a complete level order traversal of the tree and maintain a matrix of all nodes visited by level.
+2. Choose only the first element for each level and return.
+
+Time Complexity: O(n)
+Space Complexity: O(n)
+
+Optimal Approach:
+
+1. Perform DFS traversal of the tree prioritising the left subtree each time.
+2. Take the first element of each level and return.
+
+For right view, repeat the same but take the last element of the matrix or perform DFS with the right subtree prioritised.

Trees (Binary Trees)/Views of Trees/Left and Right View/Solution.cpp

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+class Solution {
+    public:
+      vector<int> leftView(Node *root) {
+          vector<int> result;
+          traverse(root, 0, result);
+          return result;
+      }
+      void traverse(Node* root, int level, vector<int>& res){
+        //BRUTE FORCE APPROACH
+          if(root == NULL) return;
+          if(res.size() <= level){
+             res.push_back({});
+          }
+          res[level].push_back(root->data);
+          traverse(root->left, level+1, res);
+          traverse(root->right, level+1, res);
+
+          //OPTIMAL APPROACH
+          if(!root) return;
+
+          if(res.size() == level){
+              res.push_back(root->data);
+          }
+
+          traverse(root->left, level+1, res);
+          traverse(root->right, level+1, res);
+      }
+  };