Find the sum of all left leaves in a given binary tree.

**Example:**

3 / \ 9 20 / \ 15 7 There are two left leaves in the binary tree, with values9and15respectively. Return24.

## Solution

### DFS Approach – Recursive

/** * Definition for a binary tree node. * public class TreeNode { * public int val; * public TreeNode left; * public TreeNode right; * public TreeNode(int x) { val = x; } * } */ public class Solution { public int SumOfLeftLeaves(TreeNode root) { return DFS(root,false); } public int DFS(TreeNode root,bool isLeft){ if(root == null){ return 0; } if(root.left == null && root.right == null && isLeft){ return root.val; } return DFS(root.left,true) + DFS(root.right,false); } }

## Iterative Approach – Using Stack

/** * Definition for a binary tree node. * public class TreeNode { * public int val; * public TreeNode left; * public TreeNode right; * public TreeNode(int x) { val = x; } * } */ public class Solution { public int SumOfLeftLeaves(TreeNode root) { int sum = 0; if(root == null) return sum; Stack<TreeNode> s = new Stack<TreeNode>(); s.Push(root); while(s.Count > 0){ TreeNode node = s.Pop(); if(node.left != null){ if(node.left.left == null && node.left.right == null){ sum += node.left.val; } else s.Push(node.left); } if(node.right != null){ if(node.right.left != null || node.right.right!=null){ s.Push(node.right); } } } return sum; } }

In both the approaches, Time Complexity and Space Complexity is O(n)