我有一个python代码，将字符串数学表达式转换为二进制树并订购树的节点，因此左孩子总是比右子小。我想按以下顺序打印二进制树。
例如，考虑数学表达式((2 * 75(/4(。buildParsetree((将字符串表达式转换为树，然后printNodeInlevels((重新排列节点，因此左子女在每个级别上都比右右子小。操作数＆LT;运营商和运营商是' '＆lt;' - '＆lt;'*'＆lt;'/'。如果树的结构就像这样
+
/
4 *
/
2 75
我想打印如下。我应该怎么做？因为数学表达式的长度始终有所不同，例如(24 * 2(，((5-1( *(2/3((，(20 - (5 4((等
Node("+") #root
.addkid(Node("*") #right child at level 1
.addkid(Node("75")) #right child at level 2
.addkid(Node("2")) #left child at level 2
)
.addkid(Node("4")) #left child at level 1
我已经制定了按级别的遍历模式打印节点的方法。如果我将该方法称为以下方法，它将打印以下内容：
pt = buildParseTree("( ( 2 * 74 ) / 4 )")
printNodesInLevels(pt)
输出：
/
4 *
2 74
这是我创建的一个函数，用于打印任何二进制树结构。
它非常通用，只需要一个启动节点(root(和一个函数(或lambda(才能获得标签和左/右子女节点：
您通常会在节点类上使用它：
printBTree(rootNode,lambda n: (n.operand, n.left, n.right) )
# assuming the Node class has a string property named operand
# and left,right properties that return a Node or None
二次方程(-b /- sqrt(b ** 2-4*a*c((/(2*a(然后可以像这样打印：
# /
# ___/ __
# +/- *
# / /
# - sqrt 2 a
#
# b -
# __/ _
# ** *
# / /
# b 2 4 *
# /
# a c
这是printbtree函数：
import functools as fn
def printBTree(node, nodeInfo=None, inverted=False, isTop=True):
# node value string and sub nodes
stringValue, leftNode, rightNode = nodeInfo(node)
stringValueWidth = len(stringValue)
# recurse to sub nodes to obtain line blocks on left and right
leftTextBlock = [] if not leftNode else printBTree(leftNode,nodeInfo,inverted,False)
rightTextBlock = [] if not rightNode else printBTree(rightNode,nodeInfo,inverted,False)
# count common and maximum number of sub node lines
commonLines = min(len(leftTextBlock),len(rightTextBlock))
subLevelLines = max(len(rightTextBlock),len(leftTextBlock))
# extend lines on shallower side to get same number of lines on both sides
leftSubLines = leftTextBlock + [""] * (subLevelLines - len(leftTextBlock))
rightSubLines = rightTextBlock + [""] * (subLevelLines - len(rightTextBlock))
# compute location of value or link bar for all left and right sub nodes
# * left node's value ends at line's width
# * right node's value starts after initial spaces
leftLineWidths = [ len(line) for line in leftSubLines ]
rightLineIndents = [ len(line)-len(line.lstrip(" ")) for line in rightSubLines ]
# top line value locations, will be used to determine position of current node & link bars
firstLeftWidth = (leftLineWidths + [0])[0]
firstRightIndent = (rightLineIndents + [0])[0]
# width of sub node link under node value (i.e. with slashes if any)
# aims to center link bars under the value if value is wide enough
#
# ValueLine: v vv vvvvvv vvvvv
# LinkLine: / / / /
#
linkSpacing = min(stringValueWidth, 2 - stringValueWidth % 2)
leftLinkBar = 1 if leftNode else 0
rightLinkBar = 1 if rightNode else 0
minLinkWidth = leftLinkBar + linkSpacing + rightLinkBar
valueOffset = (stringValueWidth - linkSpacing) // 2
# find optimal position for right side top node
# * must allow room for link bars above and between left and right top nodes
# * must not overlap lower level nodes on any given line (allow gap of minSpacing)
# * can be offset to the left if lower subNodes of right node
# have no overlap with subNodes of left node
minSpacing = 2
rightNodePosition = fn.reduce(lambda r,i: max(r,i[0] + minSpacing + firstRightIndent - i[1]),
zip(leftLineWidths,rightLineIndents[0:commonLines]),
firstLeftWidth + minLinkWidth)
# extend basic link bars (slashes) with underlines to reach left and right
# top nodes.
#
# vvvvv
# __/ __
# L R
#
linkExtraWidth = max(0, rightNodePosition - firstLeftWidth - minLinkWidth )
rightLinkExtra = linkExtraWidth // 2
leftLinkExtra = linkExtraWidth - rightLinkExtra
# build value line taking into account left indent and link bar extension (on left side)
valueIndent = max(0, firstLeftWidth + leftLinkExtra + leftLinkBar - valueOffset)
valueLine = " " * max(0,valueIndent) + stringValue
slash = "\" if inverted else "/"
backslash = "/" if inverted else "\"
uLine = "¯" if inverted else "_"
# build left side of link line
leftLink = "" if not leftNode else ( " " * firstLeftWidth + uLine * leftLinkExtra + slash)
# build right side of link line (includes blank spaces under top node value)
rightLinkOffset = linkSpacing + valueOffset * (1 - leftLinkBar)
rightLink = "" if not rightNode else ( " " * rightLinkOffset + backslash + uLine * rightLinkExtra )
# full link line (will be empty if there are no sub nodes)
linkLine = leftLink + rightLink
# will need to offset left side lines if right side sub nodes extend beyond left margin
# can happen if left subtree is shorter (in height) than right side subtree
leftIndentWidth = max(0,firstRightIndent - rightNodePosition)
leftIndent = " " * leftIndentWidth
indentedLeftLines = [ (leftIndent if line else "") + line for line in leftSubLines ]
# compute distance between left and right sublines based on their value position
# can be negative if leading spaces need to be removed from right side
mergeOffsets = [ len(line) for line in indentedLeftLines ]
mergeOffsets = [ leftIndentWidth + rightNodePosition - firstRightIndent - w for w in mergeOffsets ]
mergeOffsets = [ p if rightSubLines[i] else 0 for i,p in enumerate(mergeOffsets) ]
# combine left and right lines using computed offsets
# * indented left sub lines
# * spaces between left and right lines
# * right sub line with extra leading blanks removed.
mergedSubLines = zip(range(len(mergeOffsets)), mergeOffsets, indentedLeftLines)
mergedSubLines = [ (i,p,line + (" " * max(0,p)) ) for i,p,line in mergedSubLines ]
mergedSubLines = [ line + rightSubLines[i][max(0,-p):] for i,p,line in mergedSubLines ]
# Assemble final result combining
# * node value string
# * link line (if any)
# * merged lines from left and right sub trees (if any)
treeLines = [leftIndent + valueLine] + ( [] if not linkLine else [leftIndent + linkLine] ) + mergedSubLines
# invert final result if requested
treeLines = reversed(treeLines) if inverted and isTop else treeLines
# return intermediate tree lines or print final result
if isTop : print("n".join(treeLines))
else : return treeLines
这是使用简单的treenode类产生的输出类型的示例。
class TreeNode:
def __init__(self,rootValue):
self.value = rootValue
self.left = None
self.right = None
def addValue(self,newValue):
if newValue == self.value: return self
if newValue < self.value:
if self.left : return self.left.addValue(newValue)
self.left = TreeNode(newValue)
return self.left
if self.right : return self.right.addValue(newValue)
self.right = TreeNode(newValue)
return self.right
def printTree(self):
printBTree(self,lambda n:(str(n.value),n.left,n.right))
root = TreeNode(80)
root.addValue(50)
root.addValue(90)
root.addValue(10)
root.addValue(60)
root.addValue(30)
root.addValue(70)
root.addValue(55)
root.addValue(5)
root.addValue(35)
root.addValue(85)
root.printTree()
这会产生以下输出：
# 80
# ___/ ___
# 50 90
# __/ __ /
# 10 60 85
# / /
# 5 30 55 70
#
# 35
该函数足够通用，可以处理未存储在对象层次结构中的二进制树结构。这是一个示例，说明它如何用于从包含堆树的列表中打印：
def printHeapTree(tree, inverted=False):
def getNode(index):
left = index * 2 + 1
right = index * 2 + 2
left = left if left < len(tree) and tree[left] else None
right = right if right < len(tree) and tree[right] else None
return (str(tree[index]), left, right)
printBTree(0,getNode,inverted)
formula = ["+","4","*",None,None,"2","75"]
printHeapTree(formula)
# +
# /
# 4 *
# /
# 2 75
该功能将自动调整较宽标签的凹痕：
family = [ "Me","Paul","Rosa","Vincent","Jody","John","Kate"]
printHeapTree(family)
# Me
# ___/ ___
# Paul Rosa
# / /
# Vincent Jody John Kate
它也可以将树倒置打印(适合家庭树(：
printHeapTree(family,inverted=True)
# Vincent Jody John Kate
# / /
# Paul Rosa
# ¯¯¯ /¯¯¯
# Me
简单而粗糙的一个：
from collections import deque
def print_tree(root):
res = []
q = deque([root])
while q:
row = []
for _ in range(len(q)):
node = q.popleft()
if not node:
row.append("#")
continue
row.append(node.val)
q.append(node.left)
q.append(node.right)
res.append(row)
rows = len(res)
base = 2**(rows)
for r in range(rows):
for v in res[r]:
print("." * (base), end = "")
print(v, end = "")
print("." * (base - 1), end = "")
print("|")
base //= 2
print_tree(root)
开始，您应该阅读python的pep8代码约定，因为它说功能，属性和变量应在snake_case中。
您正在以一种迭代的方式打印，因此这意味着您无法在三角形三角形上打印它，因为您不知道基部的大小是什么(树的最低部分(，以一种迭代的方式您应该应该像具有90度角的三角形一样打印它。
，或者您可以将所有信息收集到列表或字符串和格式中，然后将其打印出来。想想头部，然后是他们之间有线条的孩子。