目录

1.示例2.术语及定义3.实现3.1 列表之列表3.2节点与引用4.二叉树的应用4.1解析树4.2树的遍历5.利用二叉堆实现优先级队列6.二叉搜索树6.1搜索树的实现7.平衡二叉搜索树（AVL树）

1.示例

树的一些属性：

(资料图片)

层次性：树是按层级构建的，越笼统就越靠近顶部，越具体则越靠近底部。一个节点的所有子节点都与另一个节点的所有子节点无关。叶子节点都是独一无二的。嵌套

2.术语及定义

节点：树的基础部分。节点的名字 → 键，附加信息 → 有效载荷。边：两个节点通过一条边相连，表示它们之间存在关系。除了根节点，其他每个结点都仅有一条入边，出边则可能有多条。根节点：树中唯一没有入边的结点。路径：由边连接的有序节点列表。子节点：一个节点通过出边与子节点相连。父节点：一个节点是其所有子节点的父节点。兄弟节点：具有同一父节点的结点 → 互称兄弟节点。子树：一个父节点及其所有后代的节点和边构成一棵子树。叶子结点：叶子节点没有子节点。层数：节点n的层数是从根节点到n的唯一路径长度。根节点的层数为0。高度：树的高度是其中节点层数的最大值。

1.定义一：树由节点及连接节点的边构成。

树的属性：

有一个根节点除根节点外，其他每个节点都与其唯一的父节点相连。从根节点到其他每个节点有且仅有一条路径。如果每个节点最多有两个子节点 → 二叉树。

2.定义二：一棵树要么为空，要么由一个根节点和零棵或多棵子树构成，子树本身也是一棵树。

每棵子树的根节点通过一条边连到父树的根节点。

3.实现

3.1 列表之列表

树的根节点是myTree[0]，左子树是myTree[1]，右子树是myTree[2]。

# 列表函数
def BinaryTree(r):
    return [r,[],[]] # 根节点r，和两个作为子节点的空列表
# 插入左子树
def insertLeft(root,newBranch):
    t = root.pop(1)
    if len(t) > 1:
        root.insert(1,[newBranch,t,[]])
    else:
        root.insert(1,[newBranch,[],[]])
    return root

## 插入右子树
def insertRight(root , newBranch):
    t = root.pop(2)
    if len(t) > 1:
        root.insert(2,[newBranch,[],t])
    else:
        root.insert(2,[newBranch,[],[]])
    return root
### 树的访问函数
def getRootVal(root):
    return root[0]
def setRootVal(root,newVal):
    root[0] = newVal
def getLeftChild(root):
    return root[1]
def getRightChild(root):
    return root[2]
r = BinaryTree(3)
insertLeft(r,4)
print(r)

3.2节点与引用

定义一个类，其中有根节点和左右子树的属性。

class BinaryTree:
    def __init__(self,rootObj):
        self.key = rootObj
        self.leftChild = None
        self.rightChild = None

    ## 插入左子节点
    def insertLeft(self,newNode):
        if self.leftChild == None:
            self.leftChild = BinaryTree(newNode)
        else:
            t = BinaryTree(newNode)
            t.left = self.leftChild
            self.leftChild = t
    ## 插入右子节点
    def insertRight(self,newNode):
        if self.rightChild == None:
            self.rightChild = BinaryTree(newNode)
        else:
            t = BinaryTree(newNode)
            t.right = self.rightChild
            self.rightChild = t
    ## 访问函数
    def getRightChild(self):
        return self.rightChild

    def getLeftChild(self):
        return self.leftChild

    def setRootVal(self,obj):
        self.key = obj
    def getRootVal(self):
        return self.key

4.二叉树的应用

4.1解析树

根据完全括号表达式构建解析树如何计算解析树中的表达式如何将解析树还原成最初的数学表达式

解析树构建器：

import operator

from pythonds.basic import Stack
from pythonds.trees import BinaryTree
def buildParseTree(fpexp):
    fplist = fpexp.split()
    pStack = Stack()
    eTree = BinaryTree("")
    pStack.push(eTree)
    currentTree = eTree

    for i in fplist:
        if i == "(":
            currentTree.insertLeft("")
            pStack.push(currentTree)
            currentTree = currentTree.getLeftChild()

        elif i not in "+-*/)":
            currentTree.setRootVal(eval(i))
            parent = pStack.pop()
            currentTree = parent
        elif i in "+-*/":
            currentTree.setRootVal(i)
            currentTree.insertRight("")
            currentTree = currentTree.getRightChild()
        elif i == ")":
            currentTree = pStack.pop()
        else:
            raise ValueError("Unkown Operator :" + i )
    return eTree

## 计算二叉解析树的递归函数
def evaluate(parseTree):
    opers = {
        "+":operator.add,"-":operator.sub,
        "*":operator.mul,"/":operator.truediv
    }
    
    leftC = parseTree.getLeftChild()
    rightC = parseTree.getRightChild()
    
    if leftC and rightC:
        fn = opers[parseTree.getRootVal()]
        return fn(evaluate(leftC),evaluate(rightC))
    else:
        return parseTree.getRootVal()

4.2树的遍历

前序遍历【根左右】中序遍历【左根右】后序遍历【左右根】

前序遍历算法实现为外部函数：

def preorder(tree):
    if tree:
        print(tree.getRootVal())
        preorder(tree.getLeftChild())
        preorder(tree.getRightChild)

前序遍历算法实现为BinaryTree类的方法

def preorder(self):
    print(self.key)
    if self.leftChild:
        self.leftChild.preorder()
    if self.rightChild:
        self.rightChild.preorder()

后序遍历函数

def postorder(tree):
    if tree != None:
        postorder(tree.getLeftChild())
        postorder(tree.getRightChild())
        print(tree.getRootVal())

中序遍历函数

def inorder(tree):
    if tree != None:
        inorder(tree.getLeftChild())
        print(tree.getRootVal())
        inorder(tree.getRightChild())

5.利用二叉堆实现优先级队列

队列一个重要的变体 → 优先级队列。和队列一样，优先级队列从头部移除元素，不过元素的逻辑顺序是由优先级决定的，优先级最高的元素在最前，最低的元素在最后。

实现优先级队列的经典方法 → 二叉堆。入队和出队操作均可达到O(logn)

最小堆【最小的元素一直在队首】最大堆【最大的元素一直在队首】 6.6.2 二叉堆的实现

结构属性：

完全二叉树：除了最底层，其他每一层的节点都是满的。且在最底层，从左往右填充节点。完全二叉树可以用一个列表直接表示。

堆的有序性：对于堆中任意元素x及其父元素p，p都不大于x。

堆操作

代码实现：

class EchaDui:
    # 新建二叉堆
    def __init__(self):
        self.heapList = [0]
        self.currentSize = 0
    
    def percUp(self,i):
        while i // 2 > 0:
            if self.heapList[i] < self.heapList[i // 2]:
                tmp = self.heapList[i // 2]
                self.heapList[i // 2] = self.heapList[i]
                self.heapList[i] = tmp
                
            i = i // 2
    # 新加元素
    def insert(self,k):
        self.heapList.append(k)
        self.currentSize = self.currentSize + 1
        self.percUp(self.currentSize)
        
    def percDown(self,i):
        while (i * 2) <= self.currentSize:
            mc = self.minChild(i)
            if self.heapList[i] > self.heapList[mc]:
                tmp = self.heapList[i]
                self.heapList[i] = self.heapList[mc]
                self.heapList[mc] = tmp
            i = mc
    def minChild(self,i):
        if i * 2 + 1 > self.currentSize:
            return i * 2
        else:
            if self.heapList[i*2] < self.heapList[i*2 + 1]:
                return i * 2
            else:
                return i * 2 + 1
    ## 从二叉堆中删除最小的元素
    def delMin(self):
        retval = self.heapList[1]
        self.heapList[1] = self.heapList[self.currentSize]
        self.currentSize = self.currentSize - 1
        self.heapList.pop()
        self.percDown(1)
        return retval
        
    ## 根据元素列表构建堆
    def builgHeap(self,alist):
        i = len(alist) // 2
        self.currentSize = len(alist)
        self.heapList = [0] + alist[:]
        while (i > 0):
            self.percDown(i)
            i = i - 1

6.二叉搜索树

6.1搜索树的实现

二叉搜索树依赖性质：小于父节点的键都在左子树中，大于父节点的键则都在右子树。

代码实现：

class BinarySearchTree:
    def __init__(self):
        self.root = None
        self.size = 0
    def length(self):
        return self.size
    def __len__(self):
        return self.size
    def __iter__(self):
        return self.root.__iter__()
    # 插入新节点
    def put(self,key,val):
        if self.root:
            self._put(key,val,self.root)
        else:
            self.root = TreeNode(key,val)
        self.size = self.size + 1

    def _put(self,key,val,currentNode):
        if key < currentNode.key:
            if currentNode.hasLeftChild():
                self._put(key,val,currentNode.leftChild)
            else:
                currentNode.leftChild = TreeNode(key,val,parent=currentNode)
        else:
            if currentNode.hasRightChild():
                self._put(key,val,currentNode.rightChild)
            else:
                currentNode.rightChild = TreeNode(key,val,parent=currentNode)
    def __setitem__(self, key, value):
        self._put(key,value)

    ## 查找键对应的值
    def get(self,key):
        if self.root:
            res = self._get(key,self.root)
            if res:
                return res.payload
            else:
                return None
        else:
            return None
    def _get(self,key,currentNode):
        if not currentNode:
            return None
        elif currentNode.key == key:
            return currentNode
        elif key < currentNode.key:
            return self._get(key,currentNode.leftChild)
        else:
            return self._get(key,currentNode.rightChild)
    def __getitem__(self, key):
        return self.get(key)

    # 检查树中是否有某个键
    def __contains__(self, key):
        if self._get(key,self.root):
            return True
        else:
            return False
    # 删除
    def delete(self,key):
        if self.size > 1:
            nodeToRemove = self._get(key,self.root)
            if nodeToRemove:
                self.remove(nodeToRemove)
                self.size = self.size - 1
            else:
                raise KeyError("Error,key not in tree")
        elif self.size == 1 and self.root.key == key:
            self.root = None
            self.size = self.size - 1
        else:
            raise KeyError("Error,key not in tree")
    def __delitem__(self, key):
        self.delete(key)

    """
        1. 待删除节点没有子节点
        2. 待删除节点只有一个子节点
        3. 待删除节点有两个子节点
    """
    # 寻找后继结点
    def findSuccessor(self):
        succ = None
        if self.hasRightChild():
            succ = self.rightChild.findMin()
        else:
            if self.parent:
                if self.isLeftChild():
                    succ = self.parent
                else:
                    self.parent.rightChild = None
                    succ = self.parent.findSuccessor()
                    self.parent.rightChild = self
        return succ

    def findMin(self):
        current = self
        while current.hasLeftChild():
            current = current.leftChild
        return current

    def spliceOut(self):
        if self.isLeaf():
            if self.isLeftChild():
                self.parent.leftChild = None
            else:
                self.parent.rightChild = None
        elif self.hasAnyChildren():
            if self.hasLeftChild():
                if self.isLeftChild():
                    self.parent.leftChild = self.leftChild
                else:
                    self.parent.rightChild = self.leftChild
                self.leftChild.parent = self.parent

            else:
                if self.isLeftChild():
                    self.parent.leftChild = self.rightChild
                else:
                    self.parent.rightChild = self.rightChild
                self.rightChild.parent = self.parent

    def remove(self,currentNode):
        if currentNode.isLeaf():
            if currentNode == currentNode.parent.leftChild:
                currentNode.parent.leftChild = None
            else:
                currentNode.parent.rightChild = None
        elif currentNode.hasBothChildren():
            succ = currentNode.findSuccessor()
            succ.spliceOut()
            currentNode.key = succ.key
            currentNode.payload = succ.payload
        else:
            if currentNode.hasLeftChild():
                if currentNode.isLeftChild():
                    currentNode.leftChild.parent = currentNode.parent
                    currentNode.parent.leftChild = currentNode.leftChild
                elif currentNode.isRightChild():
                    currentNode.leftChild.parent = currentNode.parent
                    currentNode.parent.rightChild = currentNode.leftChild
                else:
                    currentNode.replaceNodeData(currentNode.leftChild.key,
                                                currentNode.leftChild.payload,
                                                currentNode.leftChild.leftChild,
                                                currentNode.leftChild.rightChild
                                                )
            else:
                if currentNode.isLeftChild():
                    currentNode.rightChild.parent = currentNode.parent
                    currentNode.parent.leftChild = currentNode.rightChild
                elif currentNode.isRightChild():
                    currentNode.rightChild.parent = currentNode.parent
                    currentNode.parent.rightChild = currentNode.rightChild
                else:
                    currentNode.replaceNodeData(currentNode.rightChild.key,
                    currentNode.rightChild.payload,
                    currentNode.rightChild.leftChild,
                    currentNode.rightChild.rightChild                            
                                                )
    # 二叉搜索树迭代器
    def __iter__(self):
        if self:
            if self.hasLeftChild():
                for elem in self.leftChild:
                    yield elem
            yield self.key
            if self.hasRightChild():
                for elem in self.rightChild:
                    yield elem

class TreeNode:
    def __init__(self,key,val,left = None,right = None,parent = None):
        self.key = key
        self.payload = val
        self.leftChild = left
        self.rightChild = right
        self.parent = parent

    def hasLeftChild(self):
        return self.leftChild
    def hasRightChild(self):
        return self.rightChild
    def isLeftChild(self):
        return self.parent and self.parent.leftChild == self
    def isRightChild(self):
        return self.parent and self.parent.rightChild == self
    def isRoot(self):
        return not self.parent
    def isLeaf(self):
        return not (self.rightChild or self.leftChild)
    def hasAnyChildren(self):
        return self.rightChild or self.leftChild
    def replaceNodeData(self,key,value,lc,rc):
        self.key = key
        self.payload = value
        self.leftChild = lc
        self.rightChild = rc
        if self.hasLeftChild():
            self.leftChild.parent = self
        if self.hasRightChild():
            self.rightChild.parent = self

7.平衡二叉搜索树（AVL树）

实现AVL树时，要记录每个节点的平衡因子。

平衡因子 = 左右子树的高度之差

→ 保证树的平衡因子为-1，0,1，可以使得关键操作获得更好的大O性能

#from 第6章树.二叉搜索树 import TreeNode
def _put(self, key, val, currentNode):
    if key < currentNode.key:
        if currentNode.hasLeftchi1d():
            self._put(key, val, currentNode.leftChild)
        else:
            currentNode.leftChild = TreeNode(key, val,parent=currentNode)
            self.updateBalance(currentNode.leftChild)
    else:
        if currentNode.hasRightChild():
            self._put(key, val, currentNode.rightChild)
        else:
            currentNode.rightchild - TreeNode(key, val,parent=currentNode)
            self.updateBalance(currentNode.rightChild)
def updateBalance(self, node):
    if node.balanceFactor > 1 or node.balanceFactor < -1:
        self.rebalance(node)
        return
    if node.parent != None:
        if node.isLeftChild():
            node.parent.balanceFactor += 1
        elif node.isRightChild():
            node.parent.balanceFactor -= 1
        if node.parent.balanceFactor != 0:
            self.updateBalance(node.parent)
# 实现左旋
def rotateLeft (self, rotRoot) :
    newRoot = rotRoot .rightchild
    rotRoot .rightChild = newRoot.leftChild
    if newRoot . leftChild !=None :
        newRoot . leftChild. parent = rotRoot
    newRoot.parent =rotRoot.parent
    if rotRoot .isRoot( ):
        self.root = newRoot
    else:
        if rotRoot .isLeftChild():
            rotRoot.parent .leftChild = newRoot
        else:
            rotRoot.parent .rightChild = newRoot
    newRoot . leftChild = rotRoot
    rotRoot.parent = newRoot
    rotRoot. balanceFactor = rotRoot . balanceFactor + 1 - min(newRoot . balanceFactor,0)
    newRoot . balanceFactor = newRoot . balanceFactor + 1 +max(rotRoot . balanceFactor,o )

# 实现再平衡
def rebalance(self, node) :
    if node. balanceFactor < 0:
        if node .rightChild .balanceFactor > 0:
            self.rotateRight (node.rightChild)self.rotateLeft (node)
        else:
            self.rotateLeft (node)
    elif node. balanceFactor > 0 :
        if node . leftChild. balanceFactor < 0:
            self.rotateLeft (node. leftChild)
            self.rotateRight (node)
        else:
            self.rotateRight (node)
nceFactor + 1 - min(newRoot . balanceFactor,0)
    newRoot . balanceFactor = newRoot . balanceFactor + 1 +max(rotRoot . balanceFactor,o )
# 实现再平衡
def rebalance(self, node) :
    if node. balanceFactor < 0:
        if node .rightChild .balanceFactor > 0:
            self.rotateRight (node.rightChild)self.rotateLeft (node)
        else:
            self.rotateLeft (node)
    elif node. balanceFactor > 0 :
        if node . leftChild. balanceFactor < 0:
            self.rotateLeft (node. leftChild)
            self.rotateRight (node)
        else:
            self.rotateRight (node)

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关键词： 删除节点 数据结构 算法分析 节点相连