331. Verify Preorder Serialization of a Binary Tree

Description

One way to serialize a binary tree is to use preorder 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 '#'.

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 preorder, return true if it is a correct preorder traversal serialization of a binary tree.

It is guaranteed that 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".

Note: You are not allowed to reconstruct the tree.

 

Example 1:

Input: preorder = "9,3,4,#,#,1,#,#,2,#,6,#,#"
Output: true

Example 2:

Input: preorder = "1,#"
Output: false

Example 3:

Input: preorder = "9,#,#,1"
Output: false

 

Constraints:

  • 1 <= preorder.length <= 104
  • preorder consist of integers in the range [0, 100] and '#' separated by commas ','.

Solutions

Solution 1

Python Code
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class Solution:
    def isValidSerialization(self, preorder: str) -> bool:
        stk = []
        for c in preorder.split(","):
            stk.append(c)
            while len(stk) > 2 and stk[-1] == stk[-2] == "#" and stk[-3] != "#":
                stk = stk[:-3]
                stk.append("#")
        return len(stk) == 1 and stk[0] == "#"

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

C++ Code
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class Solution {
public:
    bool isValidSerialization(string preorder) {
        vector<string> stk;
        stringstream ss(preorder);
        string s;
        while (getline(ss, s, ',')) {
            stk.push_back(s);
            while (stk.size() >= 3 && stk[stk.size() - 1] == "#" && stk[stk.size() - 2] == "#" && stk[stk.size() - 3] != "#") {
                stk.pop_back();
                stk.pop_back();
                stk.pop_back();
                stk.push_back("#");
            }
        }
        return stk.size() == 1 && stk[0] == "#";
    }
};

Go Code
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func isValidSerialization(preorder string) bool {
	stk := []string{}
	for _, s := range strings.Split(preorder, ",") {
		stk = append(stk, s)
		for len(stk) >= 3 && stk[len(stk)-1] == "#" && stk[len(stk)-2] == "#" && stk[len(stk)-3] != "#" {
			stk = stk[:len(stk)-3]
			stk = append(stk, "#")
		}
	}
	return len(stk) == 1 && stk[0] == "#"
}

Solution 2

Java Code
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class Solution {
    public boolean isValidSerialization(String preorder) {
        List<String> stk = new ArrayList<>();
        for (String s : preorder.split(",")) {
            stk.add(s);
            while (stk.size() >= 3 && stk.get(stk.size() - 1).equals("#")
                && stk.get(stk.size() - 2).equals("#") && !stk.get(stk.size() - 3).equals("#")) {
                stk.remove(stk.size() - 1);
                stk.remove(stk.size() - 1);
                stk.remove(stk.size() - 1);
                stk.add("#");
            }
        }
        return stk.size() == 1 && stk.get(0).equals("#");
    }
}