如何在Python中将二叉树打印为节点结构_Python_Binary Tree_Tree Traversal

如何在Python中将二叉树打印为节点结构

python

如何在Python中将二叉树打印为节点结构,python,binary-tree,tree-traversal,Python,Binary Tree,Tree Traversal,我需要将字符串数学表达式转换为二叉树，并对树的节点进行排序，这样左边的子节点总是比右边的子节点小。我想按以下顺序打印二叉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，PrintNodeInLevel会重新排列节点，使左子级在每个级别上都小于右子级。操作数

我需要将字符串数学表达式转换为二叉树，并对树的节点进行排序，这样左边的子节点总是比右边的子节点小。我想按以下顺序打印二叉树

例如考虑数学表达式2 * 75/4。buildParseTree将字符串表达式转换为树，PrintNodeInLevel会重新排列节点，使左子级在每个级别上都小于右子级。操作数<运算符和运算符的顺序为“+”<”-“<”*“<”/”。如果树的结构是这样的

我想把它打印如下。我该怎么做？因为数学表达式的长度一直在变化，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

我已经制定了按顺序遍历模式按级别打印节点的方法。如果我按如下方式调用该方法，它将打印以下内容：

pt = buildParseTree("( ( 2 * 74 ) / 4 )")

printNodesInLevels(pt)

输出：

/ 
4 * 
2 74

首先，您应该阅读python的PEP8代码约定，因为它说函数、属性和变量应该在snake_的情况下

你是以一种迭代的方式打印的，这意味着你不能以等腰三角形打印，因为你不知道树的最底部的大小，以一种迭代的方式你应该像一个90度角的三角形一样打印

或者，您可以将所有信息收集到列表或字符串中，然后将其格式化并打印出来。首先考虑头部，然后是中间有线条的子对象。

首先，您应该阅读python的PEP8代码约定，因为它说函数、属性和变量应该在snake_的情况下

或者，您可以将所有信息收集到列表或字符串中，然后将其格式化并打印出来。考虑头部，然后是中间有线条的子对象。

我创建了一个函数，用于打印任何二叉树结构

它非常通用，只需要一个起始节点根和一个函数或lambda即可获得标签和左/右子节点：

您通常会在节点类上这样使用它：

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+/-sqrtb**2-4*A*c/2*A可以这样打印：

#         /
#     ___/ \__
#  +/-        *
#  / \       / \
# -   sqrt  2   a
#  \     \
#   b    -
#     __/ \_
#   **      *
#  /  \    / \
# b    2  4   *
#            / \
#           a   c

以下是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

下面是一个使用简单的TreeNode类生成的输出类型的示例

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()

该函数足够通用，可以处理未存储在对象层次结构中的二叉树结构。下面是如何使用它从包含堆树的列表打印的示例：

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

它还可以将树倒置打印，以适合于家谱：

printHeapTree(family,inverted=True)

# Vincent    Jody  John    Kate
#        \  /          \  /
#        Paul          Rosa
#            ¯¯¯\  /¯¯¯
#                Me

这是我创建的一个函数，用于打印任何二叉树结构