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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 = None class 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
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 = None class 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
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 = None class 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
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原文地址：http://www.cnblogs.com/scottgu/p/4108227.html
