二叉树非递归遍历——已通过LintCode

先序遍历

LintCode题目链接

class Solution:
    def preorderTraversal(self, root: TreeNode) -> List[int]:

        result = []
        stack = []

        while root or len(stack) > 0:
            while root:
                result.append(root.val)
                stack.append(root)
                root = root.left
            
            cur = stack.pop()
            root = cur.right
        
        return result

中序遍历

LintCode题目链接

class Solution:
    def inorderTraversal(self, root: TreeNode) -> List[int]:
        
        result = []
        stack = []

        while root or len(stack) > 0:
            while root:
                stack.append(root)
                root = root.left
            
            cur = stack.pop()
            result.append(cur.val)
            root = cur.right

        return result

后序遍历

LintCode题目链接

class Solution:
    def postorderTraversal(self, root: TreeNode) -> List[int]:

        result = []
        stack = []

        prev = None
        while root or len(stack) > 0:
            while root:
                stack.append(root)
                root = root.left
            
            cur = stack.pop()
            # 如果该节点存在右子树，且右子树未被访问，则将该节点重新压入栈中
            if cur.right and cur.right != prev:
                stack.append(cur)
                root = cur.right
            else:
                result.append(cur.val)
                prev = cur
                root = None
        
        return result

当做一个记录，如果大家需要解释思路的话，不妨留言，我再补上思路～