1093 Recover A Tree From Preorder Traversal
1093 Recover A Tree From Preorder Traversal
Recover a Tree From Preorder Traversal 
We run a preorder depth-first search (DFS) on the root of a binary tree.
At each node in this traversal, we output D dashes (where D is the depth of this node), then we output the value of this node. If the depth of a node is D, the depth of its immediate child is D + 1. The depth of the root node is 0.
If a node has only one child, that child is guaranteed to be the left child.
Given the output traversal of this traversal, recover the tree and return its root.
Example 1:
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**Input:** traversal = "1-2--3--4-5--6--7"
**Output:** [1,2,5,3,4,6,7]
Example 2:
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**Input:** traversal = "1-2--3---4-5--6---7"
**Output:** [1,2,5,3,null,6,null,4,null,7]
Example 3:
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**Input:** traversal = "1-401--349---90--88"
**Output:** [1,401,null,349,88,90]
Constraints:
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The number of nodes in the original tree is in the range [1, 1000].
1 <= Node.val <= 109
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class Solution:
def __init__(self):
self.index = 0
def recoverFromPreorder(self, traversal: str) -> TreeNode:
return self.helper(traversal, 0)
def helper(self, traversal, depth):
if self.index >= len(traversal):
return None
dash_count = 0
while (
self.index + dash_count < len(traversal)
and traversal[self.index + dash_count] == "-"
):
dash_count += 1
# If the number of dashes doesn't match the current depth, return null
if dash_count != depth:
return None
self.index += dash_count
# Extract the node value
value = 0
while self.index < len(traversal) and traversal[self.index].isdigit():
value = value * 10 + int(traversal[self.index])
self.index += 1
# Create the current node
node = TreeNode(value)
# Recursively build the left and right subtrees
node.left = self.helper(traversal, depth + 1)
node.right = self.helper(traversal, depth + 1)
return node
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