LeetCode Q 1028 - Recover a Tree From Preorder Traversal

We run a preorder depth first search 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 S of this traversal, recover the tree and return its root.

Example 1: Input: "1-2--3--4-5--6--7" ; Output: [1,2,5,3,4,6,7]
Example 2: Input: "1-2--3---4-5--6---7" ; Output: [1,2,5,3,null,6,null,4,null,7]
Example 3: Input: "1-401--349---90--88" ; Output: [1,401,null,349,88,90]

Note:

The number of nodes in the original tree is between 1 and 1000.
Each node will have a value between 1 and 10^9.

Solution: Using Stack

Code:

public TreeNode recoverFromPreorder(String S) {
        
  Deque<TreeNode> stack = new ArrayDeque<>();

  for (int i = 0; i < S.length(); ) {
    
    int depth = 0;
    while (i < S.length() && S.charAt(i) == '-') { depth++; i++; }

    int val = 0;
    while (i < S.length() && Character.isDigit(S.charAt(i))) {
       val = val * 10 + S.charAt(i++) - '0';
    }

    TreeNode node = new TreeNode(val);

    while (stack.size() > depth) stack.pop();
    
    if (!stack.isEmpty() && stack.peek().left == null)
      stack.peek().left = node;
    else if (!stack.isEmpty() && stack.peek().left != null)
      stack.peek().right = node;
    
    stack.push(node);
  }
    
  return stack.peekLast();
}

The key point is in the stack, we only store a branch. For example, in this tree we first store 1, 2, 4 in the stack, 1 is 2’s parent, and 2 is 4’s parent.

What should we do if we encounter another child of a node. We just pop numbers in the stack to make stack.size() == depth. For example,