LeetCode:Verify Preorder Serialization of a Binary Tree

Posted 2020-07-26 walker lee

Verify Preorder Serialization of a Binary Tree

Total Accepted: 14360 Total Submissions: 44550 Difficulty: Medium

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.

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思路一：

1.使用“栈”从左到右扫描数据；

2.当遇到数字字符，直接进栈；

3.当遇到“#”字符：

    1）如果“栈”顶不是“#”，直接入栈；

    2）如果“栈”顶也是“#”，连续再次出栈，循环直到“栈”顶非“#”；

3）“#”字符进栈。

4.最后，当stack中只剩一个“#”字符时，返回true，否则false。

java code:

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

思路二：

计算树的“入度”与“出度”。

设diff = 出度-入度 = outdegree - indegree;

1.当遇到“数字”indegree + 1,outdegree + 2（相当于diff+1），即每个非叶子结点有1个入度和2个出度；

2.当遇到“#”时，indegree + 1（相当于diffe-1）；

3.最后，当diff时返回true，否则false。

c++ code:

class Solution {
public:
    bool isValidSerialization(string preorder) {
        
        int diff = 1;
        for(int i = 0;i < preorder.size(); i++) {
            if(preorder[i] == ',') continue;
            while('0' <= preorder[i] && preorder[i] <= '9') i++;
            if(--diff < 0) return false;
            if(preorder[i] != '#') diff += 2;
        }
        return diff == 0;
    }
};