sum root to leaf numbers

Author: weiy

Tree/SumRootToLeafNumbers.py

@@ -0,0 +1,74 @@
+"""
+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.
+
+给一只二叉树，输出所有从根到叶路径的总和。
+O(n) 遍历。 字符串 -> 数字 -> 求和。
+
+测试用例：
+https://leetcode.com/problems/sum-root-to-leaf-numbers/description/
+
+24 ms beat 75%
+
+"""
+# 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 sumNumbers(self, root):
+        """
+        :type root: TreeNode
+        :rtype: int
+        """
+        
+        result = []
+        if not root:
+            return 0
+        
+        def helper(root, string):
+            if not root.left and not root.right:
+                result.append(int(string+str(root.val)))
+                return 
+            
+            if root.left:
+                helper(root.left, string+str(root.val))
+
+            if root.right:
+                helper(root.right, string+str(root.val))
+        helper(root, '')
+        return sum(result)