# Leetcode 1028. Recover a Tree From Preorder Traversal

2 min readSep 3, 2021

[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]

- we will see dashes and value, so we verify it’s level and value while scan through the string
- because it’s in-order traversal, if the stack.size bigger than level, will pop out from the stack
- if the stack.peek().left is null, will add it into stack.peek().left; else add it into stack.peek().right
- 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
- 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.