python 用迭代的方式实现二叉树的遍历

二叉树的前序遍历

思想：主要会用到数据结构 stack, 不说废话，直接上代码;

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def preorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if root is None:
            return []
        stack = []
        result = []
        stack.append(root)
        
        while(len(stack) > 0):
            node = stack.pop()
            result.append(node.val)
            if node.right:
                stack.append(node.right)
            if node.left:
                stack.append(node.left)
        return result

二叉树的中序遍历

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if root is None:
            return []
        stack = []
        result = []
        node = root
        while node or (len(stack) > 0):
            if node:
                stack.append(node)
                node = node.left
            else:
                node = stack.pop()
                result.append(node.val)
                node = node.right
        return result

二叉树的后序遍历

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def postorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if root is None:
            return []
        stack1 = []
        stack2 = []
        result = []
        stack1.append(root)
        while len(stack1) > 0:
            node = stack1.pop()
            stack2.append(node)
            if node.left:
                stack1.append(node.left)
            if node.right:
                stack1.append(node.right)
        while(len(stack2) > 0):
            top = stack2.pop()
            result.append(top.val)
        return result