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 #.
_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：
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：
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：
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;
}