Sum of left leaves

Time: O(N); Space: O(H); easy

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

Example 1:

Input: root = {TreeNode} [3,9,20,None,None,15,7]

  3
 / \
9  20
  /  \
 15   7

Output: 24

Explanation:

  • There are two left leaves in the binary tree, with values 9 and 15 respectively.

Example 2:

Input: root = {TreeNode} [1,None,2,None,3]

Output: 0

Explanatinon:

1
 \
  2
   \
    3

[6]:

class TreeNode(object):
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

[7]:

class Solution1(object):
    """
    Time: O(N)
    Space: O(H)
    """
    def sumOfLeftLeaves(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        def sumOfLeftLeavesHelper(root, is_left):
            if not root:
                return 0

            if not root.left and not root.right:
                return root.val if is_left else 0

            return sumOfLeftLeavesHelper(root.left, True) + \
                   sumOfLeftLeavesHelper(root.right, False)

        return sumOfLeftLeavesHelper(root, False)

[8]:

s = Solution1()

root = TreeNode(3)
root.left = TreeNode(9)
root.right = TreeNode(20)
root.right.left = TreeNode(15)
root.right.right = TreeNode(7)
assert s.sumOfLeftLeaves(root) == 24

root = TreeNode(1)
root.right = TreeNode(2)
root.right.right = TreeNode(3)
assert s.sumOfLeftLeaves(root) == 0