404. Sum of Left Leaves

Posted 2020-09-22 Premiumlab

https://leetcode.com/problems/sum-of-left-leaves/#/description

 

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

Example:

 

 

Hint:

 

Q: How to define if a node is a left leaf?

 

A: When the node exists and it has no right or left child.

 

See node 9 and 15 in the given example.

 

i.e. if root.left and not root.left.left and not root.left.right (== True)

 

It means the root.left is the left leaf of root. 

 

Sol1: 

 

Recursion. (I prefer recursion more than iteration.)

 

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def sumOfLeftLeaves(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        # recursion
        
        # base case: node is none
        if root is None:
            return 0
        # recursive case: left child is/isn‘t child 
        if root.left and not root.left.left and not root.left.right:
            return root.left.val + self.sumOfLeftLeaves(root.right)
        return self.sumOfLeftLeaves(root.left) + self.sumOfLeftLeaves(root.right)

 

 

Note:

 

1 When we find a left leaf, we should return the sum of its value and the right half of the root.  That‘s where recursion comes into play. 

 

2 The final return result should be the sum of all values of the left subtrees and all right subtrees. Do it recursively.

 

3 Don‘t need to use while loop to advance nodes in recursion.  

 

 

 

 

 

Sol2:

 

Recursion.

 

In this solution, we use varible to store sum of left leaf values. Easier to understand.

 

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def sumOfLeftLeaves(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        # recursion
        
        # base case: node is none
        sum = 0
        if root is None:
            return 0
        # recursive case: left child is/isn‘t child 
        if root.left and not root.left.left and not root.left.right:
            sum += root.left.val 
        sum += self.sumOfLeftLeaves(root.left) + self.sumOfLeftLeaves(root.right)
        return sum

 

 

Note:

 

1 "sum += root.left.val " stores values of left leaves in left subtress. No need to return anything.

 

"sum += self.sumOfLeftLeaves(root.left) + self.sumOfLeftLeaves(root.right)" returns all branches of the tree. It will be returned. 

 

 

Sol3:

 

Iteration using stack. （It is more efficient to use recursion!）

 

It‘s basically the mutation of traversing the tree using iteration(stack). Not recommended.  Refer here to see more ways to implement tree traversal.

 

 

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def sumOfLeftLeaves(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        # iteration
        #  traverse the whole tree using stack. If the left child is leave, add it to the result.
        
        if not root:
            return 0
        s = [root]
        res = 0
        while s:
            tmp = s.pop()
            if tmp.left:
                s.append(tmp.left)
                if not tmp.left.left and not tmp.left.right:
                    res += tmp.left.val
            if tmp.right:
                s.append(tmp.right)
        return res