二叉树关键在构建和遍历，python实现相对简单，我们在实现需要用到类，分别设置爱左右子树，根节点，然后从根进行遍历，进行判断，若为空进行树的构建，非空则返回到列表中即可,我在进行遍历时产生了一个错误思想：把根放到树中遍历，这样发现遍历时会产生一个属性缺失的错误，后来浏览数次代码没有发现这个问题，查找了一些资料才意识到，短板：类运用的不熟练。
class Node(object): def __init__(self, elem=-1, lchild=None, rchild=None): self.elem=elem self.lchild=lchild self.rchild=rchild class Tree(object): def __init__(self, root=None): self.root = root def add(self, elem): node = Node(elem) if self.root == None: self.root=node else: queue=[] queue.append(self.root) while queue: cur = queue.pop(0) if cur.lchild == None: cur.lchild=node return elif cur.rchild == None: cur.rchild = node return else: queue.append(cur.lchild) queue.append(cur.rchild) #深度遍历进行三种三种遍历 def preorder(self, root): """递归实现先序遍历""" if root == None: return print(root.elem) self.preorder(root.lchild) self.preorder(root.rchild) def inorder(self, root): """递归实现中序遍历""" if root == None: return self.inorder(root.lchild) print(root.elem) self.inorder(root.rchild) def end(self, root): """""" if root == None: return self.end(root.lchild) self.end(root.rchild) print(root.elem) #广度遍历 def rnf(self,root): if self.root == None: return queue=[] queue.append(root) while queue: cur_node=queue.pop(0) print(cur_node.elem) if cur_node.lchild != None: queue.append(cur_node.lchild) if cur_node.rchild != None: queue.append(cur_node.rchild) if __name__ == '__main__': tree=Tree() tree.add(0) tree.add(1) tree.add(2) tree.add(3) tree.add(4) tree.add(5) tree.add(6) tree.add(7) tree.add(8) tree.add(9) tree.rnf(tree.root) tree.preorder(tree.root) tree.inorder(tree.root) tree.end(tree.root)