# Leetcode 1028. Recover a Tree From Preorder Traversal

[hard]

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`

[Java]

1. we will see dashes and value, so we verify it’s level and value while scan through the string
2. because it’s in-order traversal, if the stack.size bigger than level, will pop out from the stack
3. if the stack.peek().left is null, will add it into stack.peek().left; else add it into stack.peek().right
4. to return the result, which is the root. We will need to pop all the other TreeNode in stack, until the last one -> the root
5. Trick: use three variable in two for loops. Track index, level and value at the same time. Make sure index is not larger than string.length, to avoid overflow.