Sum Root to Leaf Numbers in Python

C++Server Side ProgrammingProgramming

Suppose we have a binary tree containing digits from 0-9 only, here all root-to-leaf path could represent a number.

So if the tree is like −

This is representing two paths 21 and 23, so the output will be 21 + 23 = 44.

To solve this, we will follow these steps −

  • Create one recursive function called dfs(), this will take root, and num. initially num = 0
  • if the node is not null
    • num := num * 10 + value of node
    • if node right is not null and node left is not null, then’
      • sum := sum + num
      • num := num / 10
      • return from the function
    • dfs(right of node, num)
    • dfs(left of node, num)
    • num := num / 10
    • return from the method
  • initially sum := 0
  • call the dfs using root, and
  • return sum.

Example(Python)

Let us see the following implementation to get a better understanding −

 Live Demo

class TreeNode:
   def __init__(self, data, left = None, right = None):
      self.data = data
      self.left = left
      self.right = right
def insert(temp,data):
   que = []
   que.append(temp)
   while (len(que)):
      temp = que[0]
      que.pop(0)
      if (not temp.left):
         if data is not None:
            temp.left = TreeNode(data)
         else:
            temp.left = TreeNode(0)
         break
      else:
         que.append(temp.left)
      if (not temp.right):
         if data is not None:
            temp.right = TreeNode(data)
         else:
            temp.right = TreeNode(0)
         break
      else:
         que.append(temp.right)
def make_tree(elements):
   Tree = TreeNode(elements[0])
   for element in elements[1:]:
      insert(Tree, element)
   return Tree
class Solution(object):
   def sumNumbers(self, root):
      self.sum = 0
      self.dfs(root)
      return self.sum
   def dfs(self,node,num=0):
      if node:
         num = num*10 + node.data
         if not node.right and not node.left:
            self.sum+=num
            num/=10
            return
         self.dfs(node.right,num)
         self.dfs(node.left,num)
         num/=10
         return
ob1 = Solution()
tree = make_tree([2,1,3])
print(ob1.sumNumbers(tree))

Input

[2,1,3]

Output

44
raja
Published on 27-Feb-2020 10:55:07
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