README.md

331. Verify Preorder Serialization of a Binary Tree

One way to serialize a binary tree is to use preorder traversal. When we encounter a non-null node, we record the node's value. If it is a null node, we record using a sentinel value such as '#'.

For example, the above binary tree can be serialized to the string "9,3,4,#,#,1,#,#,2,#,6,#,#", where '#' represents a null node.

Given a string of comma-separated values preorder, return true if it is a correct preorder traversal serialization of a binary tree.

It is guaranteed that each comma-separated value in the string must be either an integer or a character '#' representing null pointer.

You may assume that the input format is always valid.

• For example, it could never contain two consecutive commas, such as "1,,3".

Example 1:

Input: preorder = "9,3,4,#,#,1,#,#,2,#,6,#,#"
Output: true

Example 2:

Input: preorder = "1,#"
Output: false

Example 3:

Input: preorder = "9,#,#,1"
Output: false

Constraints:

• 1 <= preorder.length <= 104
• preoder consist of integers in the range [0, 100] and '#' separated by commas ','.

Follow up: Find an algorithm without reconstructing the tree.

Solutions (Rust)

1. Stack

impl Solution {
    pub fn is_valid_serialization(preorder: String) -> bool {
        let mut stack = vec![];

        for x in preorder.split(',') {
            stack.push(x);
            while let Some(&[y, "#", "#"]) = stack.get(stack.len() - 3..stack.len()) {
                if y == "#" {
                    break;
                } else {
                    stack.pop();
                    stack.pop();
                    stack.pop();
                    stack.push("#");
                }
            }
        }

        &stack == &["#"]
    }
}