## Similar Problems

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# Sum Of Left Leaves - Leetcode Solution

## Companies:

LeetCode: Sum Of Left Leaves Leetcode Solution

Difficulty: Easy

Topics: binary-tree tree depth-first-search breadth-first-search

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

Example:
Input:
3
/

9 20
/

15 7

Output: 24 Explanation: There are two left leaves in the binary tree, with values 9 and 15 respectively. The sum of the left leaves is 9 + 15 = 24.

Solution: To solve this problem, we can apply recursion which is the most intuitive and easy-to-understand approach.

We can define a recursive function named `helper`

which takes three parameters:

`node`

which serves as the root of the current subtree.`is_left`

which indicates whether the current node is a left child of its parent.`sum_val`

which is the sum of all left leaves found so far.

Then, at each recursive call, we check whether the current node is a left leaf (i.e., it has a null left and right child) and if it is, we add its value to `sum_val`

.

We also recursively call the `helper`

function on the left and right subtrees of the current node, passing the appropriate arguments for `is_left`

and `sum_val`

.

The base case for the recursion is when the current node is null, in which case we simply return `sum_val`

.

Here is the code for the solution:

class Solution: def sumOfLeftLeaves(self, root: TreeNode) -> int: return self.helper(root, False, 0)

```
def helper(self, node, is_left, sum_val):
if node is None:
return sum_val
if node.left is None and node.right is None and is_left:
sum_val += node.val
sum_val = self.helper(node.left, True, sum_val)
sum_val = self.helper(node.right, False, sum_val)
return sum_val
```

Time Complexity: O(n) where n is the number of nodes in the binary tree. Space Complexity: O(h) where h is the height of the binary tree (i.e., the maximum number of function calls on the call stack at any one time). However, in the worst case where the binary tree is completely unbalanced, the height can be O(n), making the space complexity O(n).

## Sum Of Left Leaves Solution Code

`1`