1. Title
One way to serialize a binary tree is to use preorder traversal.
When we encounter a non-empty node, we can record the value of this node.
If it is an empty node, we can use a tag value record, such as #.
_9_
/ \
3 2
/ \ / \
4 1 # 6
/ \ / \ / \
# # # # # #
For example, the above binary tree can be serialized as a string "9,3,4,#,#,1,#,#,2,#,6,#,#"
, where # represents an empty node.
Given a sequence of comma-separated sequences, verify that it is the correct preorder serialization of the binary tree.
Write a feasible algorithm without reconstructing the tree .
Each character separated by a comma is either an integer or a '#' representing a null pointer.
你可以认为输入格式总是有效的,例如它永远不会包含两个连续的逗号,比如 "1,,3" 。
示例 1:
输入: "9,3,4,#,#,1,#,#,2,#,6,#,#"
输出: true
示例 2:
输入: "1,#"
输出: false
示例 3:
输入: "9,#,#,1"
输出: false
Source: LeetCode
Link: https://leetcode-cn.com/problems/verify-preorder-serialization-of-a-binary-tree
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2. Problem solving
Similar topics: LeetCode 297. Serialization and deserialization of binary trees (preorder traversal & sequence traversal)
- There is always one more empty node than the effective node
- Initial degree is 1, encounter number +1, encounter # -1
- In the process, degree must not be equal to zero, equal to zero is equivalent to the end
- Must be equal to 0 at the end
class Solution {
public:
bool isValidSerialization(string preorder) {
if(preorder[0]=='#')
return preorder.size()==1;//根节点为空
int degree = 1, i, n = preorder.size();
for(i = 0; i < n-1; ++i)
{
if(i==0 || (preorder[i]==','&& isdigit(preorder[i+1])))
degree++;//数字
else if(preorder[i] == ',' && preorder[i+1]=='#')
degree--;//空节点
if(degree==0 && i != n-2)
return false;//过程中等于0,不行
}
return degree==0;
}
};
4 ms 6.8 MB