331. Verify Preorder Serialization of a Binary Tree

最新推荐文章于 2024-08-26 07:28:39 发布

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最新推荐文章于 2024-08-26 07:28:39 发布

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分类专栏： LeetCode Stack 文章标签： LeetCode Stack Verify Preorder Seri

本文链接：https://blog.youkuaiyun.com/zshouyi/article/details/72514761

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本文介绍了一种不重建树即可验证二叉树序列化的正确性的算法。通过两种方法实现：一是利用差值计数，二是使用栈进行匹配。文章提供了具体的代码实现。

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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.

首先是根据树的性质解题，一个数字，肯定对应两个节点，所以遇到数字，count+=2，遇到#不操作。同时在每次遍历的之后，减去自身。如果最后count=0，说明符合规则，这个解题方法无视Preorder，Inorder，和Postorder。代码如下：

public boolean isValidSerialization(String preorder) {
    String[] nodes = preorder.split(",");
    int diff = 1;
    for (String node: nodes) {
        if (--diff < 0) return false;
        if (!node.equals("#")) diff += 2;
    }
    return diff == 0;
}

另一种方法，遇到两个连续的#，pop两次，第一次相当于把左节点pop出去，第二次相当于把夫节点pop出去，这是如果peek不是#，把#push进来充当上上个夫节点的右节点，如果peek还是#，接着push。代码如下：

public class Solution {
    public boolean isValidSerialization(String preorder) {
        if (preorder == null && preorder.length() == 0) {
            return false;
        }
        String[] strs = preorder.split(",");
        Stack<String> stack = new Stack<String>();
        for (String str: strs) {
            while (str.equals("#") && !stack.isEmpty() && stack.peek().equals(str)) {
                stack.pop();
                if (stack.isEmpty()) {
                    return false;
                }
                stack.pop();
            }
            stack.push(str);
        }
        return stack.size() == 1 && stack.peek().equals("#");
    }
}