[LeetCode]331. Verify Preorder Serialization of a Binary Tree

Problem Description

One way to serialize a binary tree is to use pre-order 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 #.
[https://leetcode.com/problems/verify-preorder-serialization-of-a-binary-tree/]

思路

没啥好说的，遇到俩#就弹出俩然后把栈顶的数换成#，如果栈空就说明是对的

Code

package q331;

import java.util.Stack;

public class Solution {

    public boolean isValidSerialization(String preorder) {
        Stack<String> s = new Stack<String>();
        if (preorder.length() <= 1 && !preorder.equals("#"))
            return false;
        if (preorder.charAt(0) == '#' && preorder.length() > 1)
            return false;
        String b;

        String[] str = preorder.split(",");
        s.push(str[0]);

        for (int i = 1; i < str.length; i++) {
            b = str[i];
            while (s.peek().equals("#") && b.equals("#")) {
                s.pop();
                if (s.isEmpty()) {
                    if (i != str.length - 1)
                        return false;
                    break;
                }
                if (!s.peek().equals("#")) {
                    s.pop();
                    if (s.isEmpty()) {
                        if (i != str.length - 1)
                            return false;
                        break;
                    }

                    b = s.peek();
                    s.push("#");
                }
            }
            if (b.equals(str[i]))
                s.push(b);
        }

        while (!s.isEmpty() && s.peek().equals("#"))
            s.pop();
        return s.isEmpty();
    }

//  public static void main(String[] args) {
//      Solution s = new Solution();
//      // System.out.println(s.isValidSerialization("9,3,4,#,#,1,#,#,2,#,6,#,#"));
//      // System.out.println(s.isValidSerialization("9,#,#,1"));
//      System.out.println(s.isValidSerialization("9,#,92,#,#"));
//  }

}