Verify Preorder Serialization of a Binary Tree

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最新推荐文章于 2025-07-29 19:45:28 发布

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分类专栏： leetcode 文章标签： leetcode 堆

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本文介绍了一种不重建二叉树即可验证其前序遍历序列化字符串正确性的算法。通过使用栈结构，文章详细阐述了如何处理输入字符串中的节点值和空节点标记，以及如何判断序列的有效性。

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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

这个题想到用栈来做就问题不大了，然后就是想清楚程序结束的条件。

public class Solution {
    public boolean isValidSerialization(String preorder) {
        Stack<String> stack=new Stack<String>();
        String[] strs =preorder.split(",");
        int i=0;
        while(i<strs.length){
            stack.push(strs[i]);
            while(stack.peek().equals("#")){
                stack.pop();
                if(stack.isEmpty()){
                    return i==strs.length-1;
                }
                String pop=stack.pop();
                if(pop.equals("#")){
                    stack.pop();
                    stack.push("#");
                }else{
                    stack.push(pop);
                    stack.push("#");
                    break;
                }
            }                 
            i++;
        }
        return false;
    }
}