December 2020 Leetcode Challenge
Leetcode - Pseudo-Palindromic Paths in a Binary Tree #1457
**UPDATE** 9/14/2022
The solution now gets a memory limit exception probably because of the path variable on the recursive stack. We can use a counter object instead and backtrack as we go along. Not the most optimal solution but it does pass the tests
```
c = Counter()
def isPal():
one = False
for k,v in c.items():
if v%2!=0:
if one:
return False
else:
one = True
return True
def dfs(node):
if not node:
return 0
c[node.val]+=1
if not node.left and not node.right:
if isPal():
c[node.val]-=1
return 1
else:
c[node.val]-=1
return 0
res = dfs(node.left)+dfs(node.right)
c[node.val]-=1
return res
return dfs(root)
```
