This is my attempt to make the coding experience easier for you guys so that you can easily learn what to do in today's leetcode challenge.

The goal is to find the sum of all left leaves in a binary tree. A left leaf is a leaf node that is the left child of its parent.

• I started with the root node.
• If the current node is null, returned 0.
• Check if the left child of the current node exists and if it is a leaf node (i.e., it has no left or right children).
• If the left child is a leaf node, add its value to the result and continue recursively with the right child.
• If the left child is not a leaf node, recursively call the function on both the left and right children.
• Returned the sum obtained from the left subtree and the right subtree.

Have a look at the code , still have any confusion then please let me know in the comments Keep Solving.:)

• Time complexity : $O(n)$

$n$ : number of nodes in the binary tree

• Space complexity : $O(n)$

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */

class Solution {
    public int sumOfLeftLeaves(TreeNode root) {
        if (root == null) {
            return 0;
        }
		
		// Checking if the left node is a leaf node
        if (root.left != null && root.left.left == null && root.left.right == null) {
            return root.left.val + sumOfLeftLeaves(root.right);
        }

		// Exploring the tree further
        return sumOfLeftLeaves(root.left) + sumOfLeftLeaves(root.right);
    }
}