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1028. Recover a Tree From Preorder Traversal

Difficulty Topics

Description

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:

Input: traversal = "1-2--3--4-5--6--7"
Output: [1,2,5,3,4,6,7]

Example 2:

Input: traversal = "1-2--3---4-5--6---7"
Output: [1,2,5,3,null,6,null,4,null,7]

Example 3:

Input: traversal = "1-401--349---90--88"
Output: [1,401,null,349,88,90]

 

Constraints:

  • The number of nodes in the original tree is in the range [1, 1000].
  • 1 <= Node.val <= 109

Solution

recover-a-tree-from-preorder-traversal.py
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def recoverFromPreorder(self, A: str) -> Optional[TreeNode]:
        stack = []
        n = len(A)
        index = 0

        while index < n:
            depth = 0

            while index < n and A[index] == '-':
                depth += 1
                index += 1

            val = 0
            while index < n and A[index].isdigit():
                val = val * 10 + int(A[index])
                index += 1

            while len(stack) > depth:
                stack.pop()

            node = TreeNode(val)

            if stack:
                if not stack[-1].left:
                    stack[-1].left = node
                else:
                    stack[-1].right = node

            stack.append(node)


        while len(stack) > 1:
            stack.pop()

        return stack.pop()