首页 > 代码库 > leetCode -- Binary Tree的3个水题 —— 3种非Recursive遍历
leetCode -- Binary Tree的3个水题 —— 3种非Recursive遍历
Binary Tree Preorder Traversal
Given a binary tree, return the preorder traversal of its nodes‘ values.
For example:
Given binary tree {1,#,2,3},
1 2 / 3
return [1,2,3].
Note: Recursive solution is trivial, could you do it iteratively?
‘‘‘Created on Nov 18, 2014@author: ScottGu<gu.kai.66@gmail.com, 150316990@qq.com>‘‘‘# Definition for a binary tree node# class TreeNode:# def __init__(self, x):# self.val = x# self.left = None# self.right = Noneclass Solution: # @param root, a tree node # @return a list of integers def preorderTraversal(self, root): stack=[] vals=[] if(root==None): return vals node=root stack.append(node) while(len(stack)!=0): node=stack.pop() if(node==None): continue vals.append(node.val) stack.append(node.right) stack.append(node.left) return vals
Binary Tree Inorder Traversal
Given a binary tree, return the inorder traversal of its nodes‘ values.
For example:
Given binary tree {1,#,2,3},
1 2 / 3
return [1,3,2].
Note: Recursive solution is trivial, could you do it iteratively?
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
‘‘‘Created on Nov 18, 2014@author: ScottGu<gu.kai.66@gmail.com, 150316990@qq.com>‘‘‘# Definition for a binary tree node# class TreeNode:# def __init__(self, x):# self.val = x# self.left = None# self.right = Noneclass Solution: # @param root, a tree node # @return a list of integers def inorderTraversal(self, root): stack=[] vals=[] visited={} if(root==None): return vals node=root stack.append(node) visited[node]=1 while(len(stack)!=0): if(node.left!=None and visited.has_key(node.left)==False): node=node.left stack.append(node) visited[node]=1 else: node=stack.pop() if(node==None): continue vals.append(node.val) if(node.right!=None): stack.append(node.right) node=node.right return vals
Binary Tree Postorder Traversal
Given a binary tree, return the postorder traversal of its nodes‘ values.
For example:
Given binary tree {1,#,2,3},
1 2 / 3
return [3,2,1].
Note: Recursive solution is trivial, could you do it iteratively?
‘‘‘Created on Nov 19, 2014@author: ScottGu<gu.kai.66@gmail.com, 150316990@qq.com>‘‘‘# Definition for a binary tree node# class TreeNode:# def __init__(self, x):# self.val = x# self.left = None# self.right = Noneclass Solution: # @param root, a tree node # @return a list of integers def postorderTraversal(self, root): visited={} stack=[] vals=[] if(root==None): return vals node=root stack.append(node) visited[node]=1 while(len(stack)!=0): node=stack[-1] if(node.left !=None and visited.has_key(node.left)==False): stack.append(node.left) visited[node.left]=1 continue else: if(node.right!=None and visited.has_key(node.right)==False): stack.append(node.right) visited[node.right]=1 continue node=stack.pop() if(node==None): continue vals.append(node.val) return vals
leetCode -- Binary Tree的3个水题 —— 3种非Recursive遍历
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。