Easy
Evaluate Boolean Binary Tree — Python
Full explanation · Time O(n) · Space O(h)
# Time: O(n)
# Space: O(h)
class TreeNode(object):
def __init__(self, val=0, left=None, right=None):
pass
# dfs with stack
class Solution(object):
def evaluateTree(self, root):
"""
:type root: Optional[TreeNode]
:rtype: bool
"""
INF = float("inf")
OP = {
2: lambda x, y: x or y,
3: lambda x, y: x and y
}
def iter_dfs(root):
ret = [0]
stk = [(1, (root, ret))]
while stk:
step, args = stk.pop()
if step == 1:
node, ret = args
if node.left == node.right:
ret[0] = node.val
continue
ret1, ret2 = [0], [0]
stk.append((2, (node, ret1, ret2, ret)))
stk.append((1, (node.right, ret2)))
stk.append((1, (node.left, ret1)))
elif step == 2:
node, ret1, ret2, ret = args
ret[0] = OP[node.val](ret1[0], ret2[0])
return ret[0]
return iter_dfs(root)
# Time: O(n)
# Space: O(h)
# dfs with recursion
class Solution2(object):
def evaluateTree(self, root):
"""
:type root: Optional[TreeNode]
:rtype: bool
"""
INF = float("inf")
OP = {
2: lambda x, y: x or y,
3: lambda x, y: x and y,
}
def dfs(node):
if node.left == node.right:
return node.val
return OP[node.val](dfs(node.left), dfs(node.right))
return dfs(root)