Leetcode: 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

怎么是一个合法的二叉树呢？对于叶节点，必定有两个空节点；类似于拓扑排序，可以依次消掉这些节点 - 非叶节点替换成空节点 - 一直到根节点。如果中间过程中发现某个节点的空节点数目小于两个，则是不合法的二叉树。

实现起来可以用堆栈，也可以直接字符串检查。

class Solution {
public:
bool isValidSerialization(string preorder) {
int nullCount = 0;
bool isValue = false;
for (int i = preorder.size() - 1; i >= 0; --i) {
if (preorder[i] == '#') {
++nullCount;
}
else if (preorder[i] == ',') {
if (isValue) {
if (nullCount >= 2) {
--nullCount;
}
else {
return false;
}
isValue = false;
}
}
else {
isValue = true;
}
}

return isValue && nullCount == 2 || !isValue && nullCount == 1;
}
};