二叉加权树中根节点的最大权重边和

Question

You are given a binary weighted tree find the maximum weight edge sum from root node.

Following is the tree. Maximum weight starting from root node is 9. Explanation: Node 1->Node 3 Weight = 6 and Node 3->Node-6 Weight = 3. Total Weight = 6+3 = 9

Answer 1

You can use any traversal method. A* would find the path with least number of node visits, but it requires extra memory. I would go for a simple depth-first traversal (using recursion) and just keep track of the maximum found.

Let's assume your tree is defined like this:

from collections import namedtuple

Edge = namedtuple("Edge", "weight,node")

class Node:
    def __init__(self, value, left=None, right=None):
        self.value = value
        self.left = left
        self.right = right
    

tree = Node(1, Edge(4, Node(2, Edge(1, Node(4)),
                               Edge(1, Node(5)))),
               Edge(6, Node(3, Edge(3, Node(6)),
                               Edge(0, Node(7)))))

...then you can add this method to your Node class:

    def maxweight(self):
        return max(self.left.weight + self.left.node.maxweight() if self.left else 0,
                   self.right.weight + self.right.node.maxweight() if self.right else 0)

and call it as:

print(tree.maxweight())