Verify Preorder Serialization of a Binary Tree -- LeetCode

时间:2021-10-03 14:51:35

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_
/ \ / \ / \
#
/ \ / \ / \
# # # # # #

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

思路:将字符串按照逗号分隔,然后依次读入节点。用一个栈来记录当前的节点是左孩子还是右孩子。

 class Solution {
public:
bool isValidSerialization(string preorder) {
istringstream iss(preorder);
string node;
stack<bool> leftPathTrack;
bool tracked = false;
while (getline(iss, node, ',')) {
if (tracked && leftPathTrack.empty()) return false;
tracked = true;
if (node == "#") {
//for the case the root node is '#'
if (leftPathTrack.empty()) continue;
/*
If the current node is left child, pop it and mark the next node as right.
Else, pop all the nodes in the stack until the top node is a left child,
pop it and mark the next node as right.
*/
while (!leftPathTrack.empty() && !leftPathTrack.top())
leftPathTrack.pop();
if (!leftPathTrack.empty()) {
leftPathTrack.pop();
leftPathTrack.push(false);
}
} else {
leftPathTrack.push(true);
}
}
return leftPathTrack.empty();
}
};