python 二叉树的前序、中序、后续遍历（迭代法 非递归）

二叉树的介绍：二叉树-leetcode

前序：

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

class Solution(object):
    def preorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if root is None:
            return []
        stack = [root]
        res = []
        while stack :
            temp = stack.pop()
            res.append(temp.val)
            if temp.right:
                stack.append(temp.right)
            if temp.left:
                stack.append(temp.left)
        return res

中序：
参考：二叉树的中序遍历

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

class Solution(object):
    def inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if root is None:
            return []
        res=[]
        stack = []
        temp = root
        while stack or temp:
            while temp:
                stack.append(temp)
                temp=temp.left
            temp = stack.pop()
            res.append(temp.val)
            temp=temp.right
        
        return res

后序：
参考：刷题系列 - Python用非递归实现二叉树后续遍历

想了半天没想出来非递归的思路…写出来的总是有错，然后参考了链接↑的代码……

思路：从左开始遍历并放入栈，读取没有下级节点的节点值，然后把该节点推出栈，并删除和上级节点关联；然后替换栈中最上的点，并去遍历右边子节点；直到栈为空，遍历结束。

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def postorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        traversalList = []
        nodeList = []
        # travel the node, add to node stack, if the node without any sub-node, record the val; then remove it and it's link with parnet, travel back to last one in stack.
        if root != None:
            nodeList.append(root)
            while nodeList != []:
                if nodeList[-1].left != None:
                    nodeList.append(nodeList[-1].left )
                elif nodeList[-1].right != None:
                    nodeList.append(nodeList[-1].right)
                else:
                    traversalList.append(nodeList[-1].val)
                    currentNode = nodeList.pop()
                    if nodeList != []:
                        if nodeList[-1].right == currentNode:
                            nodeList[-1].right = None
                        elif nodeList[-1].left == currentNode:
                            nodeList[-1].left = None
        return traversalList