Verify Preorder Serialization of a Binary Tree

Desicription

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:

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Input: "9,3,4,#,#,1,#,#,2,#,6,#,#"
Output: true

Example 2:

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Input: "1,#"
Output: false

Example 3:

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Input: "9,#,#,1"
Output: false

Solution

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class Solution {
public:
bool isValidSerialization(std::string preorder) {
struct node {
std::string num;
bool left = false;
bool right = false;
};

if(preorder == "#") {
return true;
}

auto stringStream = std::stringstream(preorder);

auto stack = std::stack<node>();
std::string order;
while(getline(stringStream, order, ',')) {
if(order != "#") {
stack.push(node{order, false, false});
} else {
if(stack.empty()) {
return false;
}
if(!stack.top().left) {
stack.top().left = true;
} else {
stack.top().right = true;
while(!stack.empty() && stack.top().left && stack.top().right) {
stack.pop();
if(stack.empty()) {
return !stringStream.rdbuf()->in_avail();
}
if(!stack.top().left) {
stack.top().left = true;
} else {
stack.top().right = true;
}
}
}
}
}
return stack.empty();
}
};