Leetcode 1028. 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:

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]


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


  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.

CS new grad, 6 years experience related to supply chain management. Located in Bay area