：输入字符序列，建立二叉树；按先序、中序、后序遍历二叉树（三种递归算法和中序遍历的非递归算法）的代码示例

Sure! Here is an example code in Python for constructing a binary tree and traversing it using pre-order, in-order, and post-order recursive algorithms, as well as an iterative in-order traversal algorithm:

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

def buildTree(preorder, inorder):
    if not inorder:
        return None
    
    root = TreeNode(preorder[0])
    idx = inorder.index(root.val)

    root.left = buildTree(preorder[1:idx+1], inorder[:idx])
    root.right = buildTree(preorder[idx+1:], inorder[idx+1:])
    
    return root

def preorderTraversal(root):
    if not root:
        return []
    
    return [root.val] + preorderTraversal(root.left) + preorderTraversal(root.right)

def inorderTraversal(root):
    if not root:
        return []
    
    return inorderTraversal(root.left) + [root.val] + inorderTraversal(root.right)

def postorderTraversal(root):
    if not root:
        return []
    
    return postorderTraversal(root.left) + postorderTraversal(root.right) + [root.val]

def iterativeInorderTraversal(root):
    res = []
    stack = []
    curr = root
    
    while curr or stack:
        while curr:
            stack.append(curr)
            curr = curr.left
        
        curr = stack.pop()
        res.append(curr.val)
        curr = curr.right
    
    return res

# example usage
preorder = [1, 2, 4, 5, 3, 6]
inorder = [4, 2, 5, 1, 3, 6]

root = buildTree(preorder, inorder)
print(preorderTraversal(root)) # [1, 2, 4, 5, 3, 6]
print(inorderTraversal(root)) # [4, 2, 5, 1, 3, 6]
print(postorderTraversal(root)) # [4, 5, 2, 6, 3, 1]
print(iterativeInorderTraversal(root)) # [4, 2, 5, 1, 3, 6]

I hope you find this helpful!