[LeetCode] 144. Binary Tree Preorder Traversal 二叉树的先序遍历

2021年09月15日 阅读数:1
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Given a binary tree, return the preorder traversal of its nodes' values.html

For example:
Given binary tree {1,#,2,3},node

   1
    \
     2
    /
   3

 

return [1,2,3].python

Note: Recursive solution is trivial, could you do it iteratively?app

 

树的遍历,最多见的有先序遍历,中序遍历,后序遍历和层序遍历,它们用递归实现起来都很是的简单。而题目的要求是不能使用递归求解。code

1. 用迭代和stack。2. Morris Traversal Solutionhtm

 

Python: Stack,  Time: O(n), Space: O(h) # h is the height of the treeblog

class Solution2(object):
    def preorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        result, stack = [], [(root, False)]
        while stack:
            root, is_visited = stack.pop()
            if root is None:
                continue
            if is_visited:
                result.append(root.val)
            else:
                stack.append((root.right, False))
                stack.append((root.left, False))
                stack.append((root, True))
        return result

Python: Morris, Time: O(n), Space: O(1)递归

class Solution(object):
    def preorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        result, curr = [], root
        while curr:
            if curr.left is None:
                result.append(curr.val)
                curr = curr.right
            else:
                node = curr.left
                while node.right and node.right != curr:
                    node = node.right
            
                if node.right is None:
                    result.append(curr.val)
                    node.right = curr
                    curr = curr.left
                else:
                    node.right = None
                    curr = curr.right
                
        return result  

 

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