331. Verify Preorder Serialization of a Binary Tree

Posted 2020-10-01 apanda009

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;
}