Hard
Binary Tree Maximum Path Sum — Python
Full explanation · Time O(n) · Space O(h)
# Time: O(n)
# Space: O(h), h is height of binary tree
class TreeNode(object):
def __init__(self, x):
self.val = x
self.left = None
self.right = None
class Solution(object):
# @param root, a tree node
# @return an integer
def maxPathSum(self, root):
def iter_dfs(node):
result = float("-inf")
max_sum = [0]
stk = [(1, [node, max_sum])]
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
result = max(result, node.val+max(ret1[0], 0)+max(ret2[0], 0))
ret[0] = node.val+max(ret1[0], ret2[0], 0)
return result
return iter_dfs(root)
# Time: O(n)
# Space: O(h), h is height of binary tree
class Solution2(object):
# @param root, a tree node
# @return an integer
def maxPathSum(self, root):
def dfs(node):
if not node:
return (float("-inf"), 0)
max_left, curr_left = dfs(node.left)
max_right, curr_right = dfs(node.right)
return (max(max_left, max_right, node.val+max(curr_left, 0)+max(curr_right, 0)),
node.val+max(curr_left, curr_right, 0))
return dfs(root)[0]