105 lines
2.3 KiB
Markdown
105 lines
2.3 KiB
Markdown
---
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layout: page
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title: Sum Root to Leaf Numbers
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last_solved: 2020-06-26
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category: trees
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leetcode_url: https://leetcode.com/problems/sum-root-to-leaf-numbers
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status: Attempted
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---
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Problem
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-------
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```
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Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.
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An example is the root-to-leaf path 1->2->3 which represents the number 123.
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Find the total sum of all root-to-leaf numbers.
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Note: A leaf is a node with no children.
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Example:
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Input: [1,2,3]
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1
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/ \
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2 3
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Output: 25
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Explanation:
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The root-to-leaf path 1->2 represents the number 12.
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The root-to-leaf path 1->3 represents the number 13.
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Therefore, sum = 12 + 13 = 25.
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Example 2:
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Input: [4,9,0,5,1]
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4
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/ \
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9 0
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/ \
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5 1
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Output: 1026
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Explanation:
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The root-to-leaf path 4->9->5 represents the number 495.
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The root-to-leaf path 4->9->1 represents the number 491.
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The root-to-leaf path 4->0 represents the number 40.
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Therefore, sum = 495 + 491 + 40 = 1026.
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```
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Solution
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----------
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{% highlight python %}
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def sumNumbers(root):
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if not root: return 0
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output = [] # list of strings of each path: ["12", "13"]
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def dfs(root, sofar): # sofar is list of integers as we recurse down the tree
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if root.left is None and root.right is None:
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output.append("".join(sofar+[str(root.val)]))
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if root.left:
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# sofar.append(str(root.val))
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dfs(root.left, sofar+[str(root.val)])
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if root.right:
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# sofar.append(str(root.val))
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dfs(root.right, sofar+[str(root.val)])
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dfs(root, [])
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return sum([ int(x) for x in output ])
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{% endhighlight %}
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_____________
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Anotehr solution
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{% highlight python %}
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def sumNumbers(self, root: TreeNode) -> int:
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def dfs(root, n):
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if not root:
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return 0
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n = n*10 + root.val
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if not root.left and not root.right:
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return n
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else:
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return dfs(root.left, n) + dfs(root.right, n)
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return dfs(root, 0)
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{% endhighlight %}
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