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Python Node.left方法代码示例

本文整理汇总了Python中Node.Node.left方法的典型用法代码示例。如果您正苦于以下问题:Python Node.left方法的具体用法?Python Node.left怎么用?Python Node.left使用的例子?那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在Node.Node的用法示例。


在下文中一共展示了Node.left方法的6个代码示例,这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞,您的评价将有助于系统推荐出更棒的Python代码示例。

示例1: build

# 需要导入模块: from Node import Node [as 别名]
# 或者: from Node.Node import left [as 别名]
    def build(self, array):
        if len(array) > 0:
            # counting the median, which is the key
            median = IntervalTree.get_median_point(self, array)
            # current Node
            n = Node(median)
            left = []
            right = []
            overlap = []

            for i in range(0, len(array)):
                if array[i].begin <= median <= array[i].end:
                    overlap.append(array[i])
                elif array[i].end < median:
                    left.append(array[i])
                elif array[i].begin > median:
                    right.append(array[i])

            n.overlap_begin_sort = sorted(overlap, key=operator.attrgetter("begin"))
            n.overlap_end_sort = sorted(overlap, key=operator.attrgetter("end"), reverse=True)

            if self.root is None:
                self.root = n

            n.left = self.build(left)
            n.right = self.build(right)

            return n
        else:
            return None
开发者ID:LordPatriot,项目名称:itree,代码行数:32,代码来源:IntervalTree.py

示例2: build_bst

# 需要导入模块: from Node import Node [as 别名]
# 或者: from Node.Node import left [as 别名]
def build_bst(A):
    if len(A) == 0:
        return None
    mid = len(A)/2
    root = Node(A[mid])
    root.left = build_bst(A[:mid])
    root.right = build_bst(A[mid+1:])
    return root
开发者ID:bfortuner,项目名称:boring-stuff,代码行数:10,代码来源:build_bst.py

示例3: test_searchFullTwoLevel

# 需要导入模块: from Node import Node [as 别名]
# 或者: from Node.Node import left [as 别名]
    def test_searchFullTwoLevel(self):
        node = Node()
        node.setKey(3)
        node11 = Node()
        node11.setKey(2)
        node.left = node11
        node12 = Node()
        node12.setKey(4)
        node.right = node12
        node21 = Node()
        node21.setKey(5)
        node11.left = node21
        node22 = Node()
        node22.setKey(6)
        node11.right = node22
        node23 = Node()
        node23.setKeyAndValue(7, 14)
        node12.left = node23
        node24 = Node()
        node24.setKeyAndValue(8,16)
        node12.right = node24

        self.assertEqual(node.search(7).value, 14)
开发者ID:jealous,项目名称:Katas,代码行数:25,代码来源:test_searchNode.py

示例4: insert_node

# 需要导入模块: from Node import Node [as 别名]
# 或者: from Node.Node import left [as 别名]
def insert_node(root, value):

    if root is None:
        root = Node(value)

    else:
        while root.right is not None:
            if value > root.value:
                root = root.right
            else:
                root = root.left
        if root.left is None:
            root.left = Node(value)
        else:
            root.right = Node(value)
    return root
开发者ID:raihanmasud,项目名称:ds-algo-py,代码行数:18,代码来源:node_insert_bst.py

示例5: build_exp_tree

# 需要导入模块: from Node import Node [as 别名]
# 或者: from Node.Node import left [as 别名]
def build_exp_tree(postfix):
    root = Node(postfix[-1])
    node = root
    i = len(postfix)-2
    while i >= 0:
        next = Node(postfix[i])
        if next.data in HIGHER_ORDER:
            if node.right == None:
                next.right = Node(postfix[i-1])
                next.left = Node(postfix[i-2])
                node.right = next
                i-=2
            else:
                node.left = next
                node = next
        else:
            if node.right == None:
                node.right = next
            else:
                node.left = next
                node = next
        i-=1
    return root
开发者ID:bfortuner,项目名称:boring-stuff,代码行数:25,代码来源:expression_tree.py

示例6: Node

# 需要导入模块: from Node import Node [as 别名]
# 或者: from Node.Node import left [as 别名]
        if n.right:
            q.append(n.right)
        if n.left:
            q.append(n.left)


    return False


# Tree is as shows:
#         0
#     1       2
#   3       4   5
n0 = Node(0); n1 = Node(1)
n2 = Node(2); n3 = Node(3)
n4 = Node(4); n5 = Node(5)

n0.left = n1; n0.right = n2
n1.parent = n0; n2.parent = n0

n1.left = n3; n1.right = None
n3.parent = n1

n2.left = n4; n2.right = n5
n4.parent = n2; n5.parent = n2

print "The first common ancestor is: %s" % firstCommonAncestor(n0, n4, n5)
print "The first common ancestor is: %s" % firstCommonAncestor(n0, n3, n4)
print "The first common ancestor is: %s" % firstCommonAncestor(n0, n3, n1)
print "The first common ancestor is: %s" % firstCommonAncestor(n0, n1, Node(0))
开发者ID:HectorRamosJunior,项目名称:Algorithms-and-Data-Structures,代码行数:32,代码来源:8Review2.py


注:本文中的Node.Node.left方法示例由纯净天空整理自Github/MSDocs等开源代码及文档管理平台,相关代码片段筛选自各路编程大神贡献的开源项目,源码版权归原作者所有,传播和使用请参考对应项目的License;未经允许,请勿转载。