Python programmers interview book --- problem-solving algorithm Summary: Chapter 3.4 binary largest sub-tree and how seek a binary tree

# -*- coding: utf-8 -*-

'''
Python程序员面试算法宝典---解题总结: 第三章 二叉树 3.4 如何求一棵二叉树的最大子树和

题目:
给定一棵二叉树，它的每个节点都是正整数或负整数，如何找到一棵子树，使得它所有节点的
和最大

分析:
举例如下，假设有一棵二叉树如下
                    -3
            2               -1
        -1      4        4      5
那么求最大子树和，必然要求这颗最大子树的根节点。
要求一棵子树的结点和，就需要先求出该子树的左孩子为根节点的子树的结点和，
和该子树的右孩子为根节点的子树的结点和。
这种情况是典型的递归做法。
那么在求解的过程中需要记录一个maxSum和maxRoot来保存
求解出leftMax, rightMax，然后计算以当前节点node为根节点的子树和
leftMax + rightMax + node.data
关键:
1
求解出以root为根节点的子树的所有节点和
注意更新的最大值如果用局部变量会有问题，
所以用类中的全局变量
注意: 整颗树也是作为一颗子树

2 代码
class SubTree(object):
    def __init__(self):
        self.maxSum = float('-inf')
        self.maxRoot = None

    def maxSumTree(self, root):
        if not root:
            return 0
        maxLeft = self.maxSumTree(root.left)
        maxRight = self.maxSumTree(root.right)
        sum = maxLeft + maxRight + root.data
        if sum > self.maxSum:
            self.maxSum = sum
            self.maxRoot = root
        # 返回以当前结点为根节点的子树的所有结点和
        return sum

参考:
Python程序员面试算法宝典
'''

class BinaryTreeNode(object):

    def __init__(self, data, left=None, right=None):
        self.data = data
        self.left = left
        self.right = right


def buildTree(data):
    if not data:
        return
    nodes = list()
    for value in data:
        node = BinaryTreeNode(value)
        nodes.append(node)
    # 构建二叉树
    length = len(data)
    for i in range(length / 2):
        node = nodes[i]
        if not node:
            continue
        if 2*i + 1 < length:
            node.left = nodes[2*i + 1]
        if 2*i + 2 < length:
            node.right = nodes[2*i + 2]
    return nodes[0]

# def maxSumTree(root, maxSum, maxRoot):
#     if not root:
#         return
#     maxLeft = maxSumTree(root.left, maxSum, maxRoot)
#     maxRight = maxSumTree(root.right, maxSum, maxRoot)
#     if maxLeft is None and maxRight is None:
#         return root.data
#     elif maxRight is None:
#         if maxLeft > maxSum:
#             maxSum = maxLeft
#             maxRoot = root.left
#     elif maxLeft is None:
#         if maxRight > maxSum:
#             maxSum = maxRight
#             maxRoot = root.right
#     else:
#         if maxLeft > maxRight:
#             if maxLeft > maxSum:
#                 maxSum = maxLeft
#                 maxRoot = root.left
#         else:
#             if maxRight > maxSum:
#                 maxSum = maxRight
#                 maxRoot = root.right

'''
求解出以root为根节点的子树的所有节点和
注意更新的最大值如果用局部变量会有问题，
所以用类中的全局变量
注意: 整颗树也是作为一颗子树
'''
class SubTree(object):
    def __init__(self):
        self.maxSum = float('-inf')
        self.maxRoot = None

    def maxSumTree(self, root):
        if not root:
            return 0
        maxLeft = self.maxSumTree(root.left)
        maxRight = self.maxSumTree(root.right)
        sum = maxLeft + maxRight + root.data
        if sum > self.maxSum:
            self.maxSum = sum
            self.maxRoot = root
        # 返回以当前结点为根节点的子树的所有结点和
        return sum

def process():
    data = [-3, 2, -1, -1, 4, 4, 5]
    root = buildTree(data)
    subTree = SubTree()
    subTree.maxSumTree(root)
    result = subTree.maxSum
    print result
    print subTree.maxRoot.data

if __name__ == "__main__":
    process()