LintCode-二叉树的前、中、后序遍历-递归

描述

给出一棵二叉树，返回其节点值的前、中、后序遍历。

样例

给出一棵二叉树 {1,#,2,3},

1

2
/
3
返回 [3,2,1]

挑战

你能使用非递归实现么？

代码（递归）

前序遍历

"""
Definition of TreeNode:
class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left, self.right = None, None
"""

class Solution:
    """
    @param root: A Tree
    @return: Preorder in ArrayList which contains node values.
    """
    def preorderTraversal(self, root):
        # write your code here
        result = []
        if root is None:
            return result
        queue = [root]
        while queue:
            root = queue.pop()
            result.append(root.val)
            if root.right is not None:
                queue.append(root.right)
            if root.left is not None:
                queue.append(root.left)
        return result

中序遍历
第一种写法

"""
Definition of TreeNode:
class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left, self.right = None, None
"""

class Solution:
    """
    @param root: A Tree
    @return: Inorder in ArrayList which contains node values.
    """
    def inorderTraversal(self, root):
        # write your code here
        if root is None:
            return []
        stack = [root]
        result = []
        while stack:
            new_root = stack.pop()
            if new_root.left is not None:
                stack.append(new_root)
                stack.append(new_root.left)
                continue
            else:
                result.append(new_root.val)
                while new_root.right is None:
                    try:
                        new_root = stack.pop()
                    except IndexError:
                        return result
                    result.append(new_root.val)
                stack.append(new_root.right)
        return result
                    

第二种写法

"""
Definition of TreeNode:
class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left, self.right = None, None
"""

class Solution:
    """
    @param root: A Tree
    @return: Inorder in ArrayList which contains node values.
    """
    def inorderTraversal(self, root):
        # write your code here
        if root is None:
            return []
        stack = [root]
        result = []
        while stack:
            new_root = stack[-1]
            if new_root.left is None and new_root.right is None:
                result.append(new_root.val)
                stack.pop()
                while stack:
                    new_root = stack.pop()
                    result.append(new_root.val)
                    if new_root.right is not None:
                        stack.append(new_root.right)
                        break
            else:
                if new_root.left is not None:
                    stack.append(new_root.left)
                else:
                    if new_root.right is not None:
                        stack.pop()
                        result.append(new_root.val)
                        stack.append(new_root.right)
        return result
                    

后序遍历

"""
Definition of TreeNode:
class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left, self.right = None, None
"""

class Solution:
    """
    @param root: A Tree
    @return: Postorder in ArrayList which contains node values.
    """
    def postorderTraversal(self, root):
        # write your code here
        result = []
        if root is None:
            return result                                       
        stack = [root]
        already = []
        while stack:
            new_root = stack.pop()
            if new_root in already:
                result.append(new_root.val)
                continue
            if new_root.left is None and new_root.right is None:
                result.append(new_root.val)
            else:
                stack.append(new_root)
                already.append(new_root)
            if new_root.right is not None:
                stack.append(new_root.right)
            if new_root.left is not None:
                stack.append(new_root.left)
        return result