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Construct String from Binary Tree in Python
Suppose we have a binary tree we have to make a string consists of parenthesis and integers from a binary tree with the preorder traversing way. A null node will be represented by empty parenthesis pair "()". And we need to omit all the empty parenthesis pairs that don't affect the one-to-one mapping relationship between the string and the original binary tree.
So, if the input is like
then the output will be 5(6()(8))(7)
To solve this, we will follow these steps −
- ans := blank string
- Define a function pot(). This will take node, s
- if is null, then
- return blank string
- if if left or right of node is null, then
- return node.data
- ss := node.data
- if node.left is not null, then
- ss := ss concatenate '(' concatenate pot(node.left, blank string) concatenate ')'
- otherwise,
- ss := ss concatenate '()'
- if node.right is not null, then
- ss := ss concatenate '(' concatenate pot(node.right, blank string) concatenate ')'
- return ss
- From the main method do the following −
- return pot(t, blank string)
Let us see the following implementation to get better understanding −
Example
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:
def tree2str(self, t: TreeNode):
ans = ''
def pot(node, s):
if node == None or node.data == 0:
return ''
if node.left == node.right == None:
return str(node.data)
ss = str(node.data)
if node.left != None:
ss = ss + '(' + pot(node.left, '') + ')'
else:
ss = ss + '()'
if node.right != None:
ss = ss + '(' + pot(node.right, '') + ')'
return ss
return pot(t, '')
ob = Solution()
root = make_tree([5,6,7,None,8])
print(ob.tree2str(root))Input
[5,6,7,None,8]
Output
5(6()(8))(7)
Updated on: 04-Jul-2020
673 Views
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