Problem

Given the root of a binary tree, imagine yourself standing on the left side of it. Return the values of the nodes you can see ordered from top to bottom.

What is Left View of a Binary Tree?

When just look at the tree from the left side , all the nodes you can see will be the left view of the tree.

Examples

Example 1:

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Input:
root = [1,2,3,4,5,6,7, null, 8]
Output:
 [1, 2, 4, 8]

Solution

This problem is similar to Binary Tree Right Side View. The left view of a binary tree can be obtained by traversing the tree level-by-level (using a level order traversal) and collecting the first node at each level.

Method 1 - Use the Level Change Indication and Recursing on Left First

  • Traverse the tree from left to right
  • Print the first node you encounter
  • Take two variables , currentLevel=0 and nextLevel=1
  • As soon as you change level , change the currentLevel = nextLevel
  • Print only when current level<nextLevel so this way you will print only the first element
  • For rest of the nodes on the the level currentLevel and nextLevel are equal so it wont print

Code

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class Solution {
    public List<Integer> leftView(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        if (root == null) {
            return result;
        }
        
        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        
        while (!queue.isEmpty()) {
            int levelSize = queue.size();

            for (int i = 0; i < levelSize; i++) {
                TreeNode currentNode = queue.poll();
                
                if (i == 0) {
                    result.add(currentNode.val);
                }
                
                if (currentNode.left != null) {
                    queue.add(currentNode.left);
                }
                
                if (currentNode.right != null) {
                    queue.add(currentNode.right);
                }
            }
        }
        
        return result;
    }
}
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class Solution:
    def leftView(self, root: TreeNode) -> List[int]:
        result = []
        if not root:
            return result
        
        queue = deque([root])
        
        while queue:
            level_size = len(queue)
            
            for i in range(level_size):
                node = queue.popleft()
                
                if i == 0:
                    result.append(node.val)
                
                if node.left:
                    queue.append(node.left)
                
                if node.right:
                    queue.append(node.right)
        
        return result

Complexity

  • ⏰ Time complexity: O(n) - where n is the number of nodes in the binary tree. Each node is visited once.
  • 🧺 Space complexity: O(n) - due to the storage of nodes in the queue used for level order traversal.