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binary_tree_path_sum.py
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64 lines (56 loc) · 1.85 KB
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#!/usr/bin/env python
'''
Leetcode: Binary Tree Maximum Path Sum
Given a binary tree, find the maximum path sum.
The path may start and end at any node in the tree.
For example: Given the below binary tree,
1
/ \
2 3
Return 6.
2-1-3
'''
from __future__ import division
import random
from BinaryTree import Node, SampleTreeRoot
# L(node) = max sum path in node's left sub-tree
# R(node) = max sum path in node's right sub-tree
# L(node) = node.val + max{L(node.left), R(node.left)}
# R(node) = node.val + max{L(node.right), R(node.right)}
# max{L(node)+R(node) for any node in the tree}
def subtree_maxsum_path(node, is_left=True, Lpath={}, Rpath={}):
if is_left:
# left sub-tree
if not node.left:
Lpath[node.id] = 0
return 0
else:
Lpath[node.id] = node.value + max(
subtree_maxsum_path(node.left, True, Lpath, Rpath),
subtree_maxsum_path(node.left, False, Lpath, Rpath)
)
return Lpath[node.id]
else:
# right sub-tree
if not node.right:
Rpath[node.id] = 0
return 0
else:
Rpath[node.id] = node.value + max(
subtree_maxsum_path(node.right, True, Lpath, Rpath),
subtree_maxsum_path(node.right, False, Lpath, Rpath)
)
return Rpath[node.id]
def maxsum_path(root):
Lpath = {}
Rpath = {}
subtree_maxsum_path(root, True, Lpath, Rpath)
subtree_maxsum_path(root, False, Lpath, Rpath)
print 'Left-path:', Lpath
print 'Right-path:', Rpath
path2sum = dict((i, Lpath[i]+Rpath[i]) for i in Lpath.keys())
i = max(path2sum, key=path2sum.get)
print 'The path going through node', i, 'with max sum', path2sum[i]
return path2sum[i]
if __name__ == '__main__':
print maxsum_path(SampleTreeRoot)