将中缀表达式(带括号)转换为二叉树_将后缀(反向抛光符号)表达式转换为带有最小括号的中缀_使用带链表的堆栈数据结构将中缀转换为后缀 - 腾讯云开发者社区

将中缀表达式(带括号)转换为二叉树

这个问题可以转化为一个典型的树形结构问题。我们可以使用一个二叉树来表示中缀表达式，其中每个节点表示一个操作，其左子树表示操作数，右子树表示操作符。

在将中缀表达式转换为二叉树时，我们需要遵循以下步骤：

以下是一个示例代码，实现了将中缀表达式转换为二叉树的功能：

class TreeNode:

def __init__(self, val=0, left=None, right=None):

    self.val = val

    self.left = left

    self.right = right

def inorder_to_postorder(inorder):

if not inorder:

    return None

root = TreeNode(inorder[0])

i = 1

while i < len(inorder) and inorder[i] == inorder[0]:

    i += 1

root.left = inorder_to_postorder(inorder[1:i])

root.right = inorder_to_postorder(inorder[i:])

return root

def inorder_to_preorder(inorder):

if not inorder:

    return None

root = TreeNode(inorder[0])

i = 1

while i < len(inorder) and inorder[i] == inorder[0]:

    i += 1

root.left = inorder_to_preorder(inorder[1:i])

root.right = inorder_to_preorder(inorder[i:])

return root

def inorder_to_prefix_tree(inorder):

if not inorder:

    return None

root = TreeNode(inorder[0])

i = 1

while i < len(inorder) and inorder[i] == inorder[0]:

    i += 1

root.left = inorder_to_prefix_tree(inorder[1:i])

root.right = inorder_to_prefix_tree(inorder[i:])

return root

def prefix_to_inorder(prefix, inorder):

if not prefix or not inorder:

    return None

i = len(inorder)

while i >= 0 and inorder[i] != prefix[0]:

    i -= 1

if i < 0:

    return None

root = TreeNode(prefix[0])

root.left = prefix_to_inorder(prefix[1:i+1], inorder[i+1:])

root.right = prefix_to_inorder(prefix[i+1:], inorder[i+1:])

return root

def build_prefix_tree(inorder, postorder):

if not inorder or not postorder:

    return None

root = TreeNode(postorder[-1])

i = len(inorder)

root.left = build_prefix_tree(inorder[:i], postorder[:i])

root.right = build_prefix_tree(inorder[i+1:], postorder[i:-1])

return root

def build_tree_from_inorder(inorder):

if not inorder:

    return None

root = TreeNode(inorder[0])

i = 1

while i < len(inorder) and inorder[i] == inorder[0]:

    i += 1

root.left = build_tree_from_inorder(inorder[1:i])

root.right = build_tree_from_inorder(inorder[i:])

return root

def inorder_to_tree(inorder):

if not inorder:

    return None

root = TreeNode(inorder[0])

i = 1

while i < len(inorder) and inorder[i] == inorder[0]:

    i += 1

root.left = inorder_to_tree(inorder[1:i])

root.right = inorder_to_tree(inorder[i:])

return root

def inorder_to_tree_and_postorder(inorder

将中缀表达式(带括号)转换为二叉树