Verify-Preorder-Serialization-of-a-Binary-Tree

LeetCode
,

第104天。

今天的题目是331. Verify Preorder Serialization of a Binary Tree:

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 #.

1
2
3
4
5
6
7
     _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

虽然题目很长，但是理解起来并不难，就是给你一串字符串表示一棵二叉树，用,分隔节点的值，用#表示空指针，然后问你这个字符串能不能还原出来一棵二叉树（在不建树的情况下）,其实和建树很像，都是递归的去做：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
bool isValidSerialization(string preorder) {
    int beg = 0;
    return isValidSerialization(preorder,beg) && !next(preorder, beg);
}
bool isValidSerialization(string preorder, int &beg) {
    if (beg >= preorder.size()) return false;
    if (preorder[beg] == '#') return true;
    return next(preorder, beg) && isValidSerialization(preorder, beg) &&
        next(preorder, beg) && isValidSerialization(preorder, beg);
}
bool next(string &preorder, int &beg) {
    while(beg < preorder.size() && preorder[beg] != ',') beg++;
    beg++;
    return beg < preorder.size();
}

然后是dicuss中的迭代版本：

1
2
3
4
5
6
7
8
9
10
11
12
13
bool isValidSerialization(string preorder) {
    if (preorder.empty()) return false;
    preorder+=',';
    int sz=preorder.size(),idx=0;
    int capacity=1;
    for (idx=0;idx<sz;idx++){
        if (preorder[idx]!=',') continue;
        capacity--;
        if (capacity<0) return false;
        if (preorder[idx-1]!='#') capacity+=2;
    }
    return capacity==0;
}

Update at 2020-04-02

补充一个自己的写法：

1
2
3
4
5
6
7
8
9
10
bool isValidSerialization(string preorder) {
    int i, c, size; 
    for(i = 0, c = 1, size = preorder.size(); i < size && c; i++){
        if (preorder[i] == '#') c--;
        else c++; // c--; c+=2;
        // move next node
        while(i < size && preorder[i]!=',') i++;
    }
    return c == 0 && i >= size;
}