86 lines
3.0 KiB
C++
86 lines
3.0 KiB
C++
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// Source : https://leetcode.com/problems/verify-preorder-serialization-of-a-binary-tree/
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// Author : Hao Chen
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// Date : 2017-01-06
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/***************************************************************************************
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*
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* One way to serialize a binary tree is to use pre-order traversal. When we encounter
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* a non-null node, we record the node's value. If it is a null node, we record using a
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* sentinel value such as #.
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*
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* _9_
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* / \
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* 3 2
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* / \ / \
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* 4 1 # 6
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* / \ / \ / \
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* # # # # # #
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*
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* For example, the above binary tree can be serialized to the string
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* "9,3,4,#,#,1,#,#,2,#,6,#,#", where # represents a null node.
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*
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* Given a string of comma separated values, verify whether it is a correct preorder
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* traversal serialization of a binary tree. Find an algorithm without reconstructing
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* the tree.
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*
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* Each comma separated value in the string must be either an integer or a character
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* '#' representing null pointer.
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*
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* You may assume that the input format is always valid, for example it could never
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* contain two consecutive commas such as "1,,3".
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*
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* Example 1:
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* "9,3,4,#,#,1,#,#,2,#,6,#,#"
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* Return true
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* Example 2:
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* "1,#"
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* Return false
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* Example 3:
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* "9,#,#,1"
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* Return false
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*
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* Credits:Special thanks to @dietpepsi for adding this problem and creating all test
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* cases.
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***************************************************************************************/
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class Solution {
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public:
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// we know the following facts:
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// 1) if we met a non-null node, then this node will generate two child node.
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// 2) if we met a null node, then this node will generate zero child node.
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//
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// so the idea is,
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// 1) we can have a counter to calculate how many node we are going to expect
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// 2) once we have the expected node, then we can decrease the counter.
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// 3) finally, we will check the counter is zero or not.
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//
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// the algorithm as below:
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// 1) when we meet a non-null node, just simply do `counter++`. because:
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// 1.1) we will expect 2 more node after, then we do `counter += 2`.
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// 1.2) but the current node also meet the expection of parent node , so, it need remove 1 in counter.
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// finally, the `counter = counbter + 2 -1`
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// 2) when we meet a null node, just simply do `counter--`.
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bool isValidSerialization(string preorder) {
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vector<string> list;
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split(preorder, ',', list);
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//we initailize the counter as 1,
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//because we expect at lease 1 node in the tree.
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int node_expected = 1;
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for (auto node : list) {
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if (node_expected == 0) return false;
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node == "#" ? node_expected-- : node_expected++;
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}
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return node_expected == 0;
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}
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void split(const string &s, char delim, vector<string> &elems) {
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stringstream ss(s);
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string item;
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while (getline(ss, item, delim)) {
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elems.push_back(item);
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}
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}
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};
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