leetcode_gym/_problems/sumRootToLeaf.md

layout, title, last_solved, category, leetcode_url, status

layout	title	last_solved	category	leetcode_url	status
page	Sum Root to Leaf Numbers	2020-06-26	trees	https://leetcode.com/problems/sum-root-to-leaf-numbers	Attempted

Problem

Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.

An example is the root-to-leaf path 1->2->3 which represents the number 123.

Find the total sum of all root-to-leaf numbers.

Note: A leaf is a node with no children.

Example:

Input: [1,2,3]
    1
   / \
  2   3
Output: 25
Explanation:
The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.
Therefore, sum = 12 + 13 = 25.
Example 2:

Input: [4,9,0,5,1]
    4
   / \
  9   0
 / \
5   1
Output: 1026
Explanation:
The root-to-leaf path 4->9->5 represents the number 495.
The root-to-leaf path 4->9->1 represents the number 491.
The root-to-leaf path 4->0 represents the number 40.
Therefore, sum = 495 + 491 + 40 = 1026.

Solution

{% highlight python %}

def sumNumbers(root):

if not root: return 0

output = []     # list of strings of each path: ["12", "13"]

def dfs(root, sofar):   # sofar is list of integers as we recurse down the tree
    
    if root.left is None and root.right is None:
        output.append("".join(sofar+[str(root.val)]))
        
    if root.left:
        # sofar.append(str(root.val))
        dfs(root.left, sofar+[str(root.val)])
    
    if root.right:
        # sofar.append(str(root.val))
        dfs(root.right, sofar+[str(root.val)])

dfs(root, [])

return sum([ int(x) for x in output ])

{% endhighlight %}

---

Anotehr solution

{% highlight python %}

def sumNumbers(self, root: TreeNode) -> int:

def dfs(root, n):
    if not root:
        return 0
    n = n*10 + root.val
    if not root.left and not root.right:
        return n
    else:
        return dfs(root.left, n) + dfs(root.right, n)
return dfs(root, 0)

{% endhighlight %}