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lowest-common-ancestor-of-a-binary-tree-ii.py
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lowest-common-ancestor-of-a-binary-tree-ii.py
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# Time: O(n)
# Space: O(h)
# Definition for a binary tree node.
class TreeNode(object):
def __init__(self, x):
pass
class Solution(object):
def lowestCommonAncestor(self, root, p, q):
"""
:type root: TreeNode
:type p: TreeNode
:type q: TreeNode
:rtype: TreeNode
"""
def iter_dfs(node, p, q):
result = None
stk = [(1, (node, [0]))]
while stk:
step, params = stk.pop()
if step == 1:
node, ret = params
if not node:
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 = params
curr = int(node == p or node == q)
if curr+ret1[0]+ret2[0] == 2 and not result:
result = node
ret[0] = curr+ret1[0]+ret2[0]
return result
return iter_dfs(root, p, q)
# Time: O(n)
# Space: O(h)
class Solution2(object):
def lowestCommonAncestor(self, root, p, q):
"""
:type root: TreeNode
:type p: TreeNode
:type q: TreeNode
:rtype: TreeNode
"""
def dfs(node, p, q, result):
if not node:
return 0
left = dfs(node.left, p, q, result)
right = dfs(node.right, p, q, result)
curr = int(node == p or node == q)
if curr+left+right == 2 and not result[0]:
result[0] = node
return curr+left+right
result = [0]
dfs(root, p, q, result)
return result[0]