Code & Fun

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 #.

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     _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:

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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:

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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;
}

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