331. Verify Preorder Serialization of a Binary Tree

最新推荐文章于 2024-08-26 07:28:39 发布

SHURamos

最新推荐文章于 2024-08-26 07:28:39 发布

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分类专栏： Leetcode

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本文介绍了一种不重建树即可验证字符串是否为正确二叉树前序遍历序列的方法。通过对输入字符串进行分析，利用栈来判断序列的有效性。

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

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

public class Solution {
    public boolean isValidSerialization(String preorder) {
        String[] a = preorder.split(",");
		Stack<String> data = new Stack<>();
		for (int i = 0; i < a.length; i++) {
			String temp = a[i];
			if (data.size() == 1 && data.peek().equals("#"))
				return false;
			if (temp.equals("#")) {
				while (!data.isEmpty() && data.peek().equals("#")) {
					data.pop();
					if (!data.peek().equals("#"))
						data.pop();
					else
						return false;
				}
			}
			data.push(temp);
		}
		return data.size() == 1 && data.pop().equals("#");
    }
}