331. Verify Preorder Serialization of a Binary Tree
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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.
找规律?
我们通过举一些正确的例子,比如"9,3,4,#,#,1,#,#,2,#,6,#,#" 或者"9,3,4,#,#,1,#,#,2,#,6,#,#"等等,可以观察出如下两个规律:
1. 数字的个数总是比#号少一个
2. 最后一个一定是#号
那么我们加入先不考虑最后一个#号,那么此时数字和#号的个数应该相同,如果我们初始化一个为0的计数器,遇到数字,计数器加1,遇到#号,计数器减1,那么到最后计数器应该还是0。下面我们再来看两个返回False的例子,"#,7,6,9,#,#,#"和"7,2,#,2,#,#,#,6,#",那么通过这两个反例我们可以看出,如果根节点为空的话,后面不能再有节点,而且不能有三个连续的#号出现。所以我们再加减计数器的时候,如果遇到#号,且此时计数器已经为0了,再减就成负数了,就直接返回False了,因为正确的序列里,任何一个位置i,在[0, i]范围内的#号数都不大于数字的个数的。当循环完成后,我们检测计数器是否为0的同时还要看看最后一个字符是不是#号。
public class Solution { public boolean isValidSerialization(String preorder) { if (preorder == null || preorder.length() == 0) return false; String[] strs = preorder.split(","); int depth = 0; int i = 0; while (i < strs.length - 1) { if (strs[i++].equals("#")) { if (depth == 0) return false; else depth--; } else depth++; } if (depth != 0) return false; return strs[strs.length - 1].equals("#"); } }
出度入度
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; }
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