331. Verify Preorder Serialization of a Binary Tree-优快云博客

本文介绍了一种不重建树即可验证字符串是否为有效二叉树前序遍历序列的方法。通过跟踪允许出现空节点的位置数量，算法可以高效判断序列的有效性。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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 #.

     _9_
    /   \
   3     2
  / \   / \
 4   1  #  6
/ \ / \   / \
# # # #   # #

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, verify whether it is a correct preorder traversal serialization of a binary tree. Find an algorithm without reconstructing the tree.

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:
"9,3,4,#,#,1,#,#,2,#,6,#,#"
Return true

Example 2:
"1,#"
Return false

Example 3:
"9,#,#,1"
Return false

The idea we need to find how many null node do we allow at the time.

At first we allow only one null node since the root could be null.

So we set count = 1;

when a null node comes count -= 1;

when a not-null node comes count+=1;

At any time count should not be less than 0;

Only at the end count should be 0, and count must be 0.

Code:

public class Solution {
    public boolean isValidSerialization(String preorder) {
        String[] nodes = preorder.split(",");
        if(nodes.length == 0 || (nodes.length == 1 && nodes[0].equals("#")))
            return true;
        int count = 1;
        for(int i = 0 ; i < nodes.length; i++){
            if(!nodes[i].equals("#")){
                count += 1;
            } else{
                count -= 1;
            }
            if(count < 0) return false;
            if(count == 0 && i != nodes.length -1) return false;
        }
        return count == 0;
    }
}