LEETCODE —— Binary Tree的3 题 —

Binary Tree Preorder Traversal

Given a binary tree, return the preorder traversal of its nodes' values.

For example:
Given binary tree {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

Binary Tree Inorder Traversal

Given a binary tree, return the inorder traversal of its nodes' values.

For example:
Given binary tree {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

Binary Tree Postorder Traversal

Given a binary tree, return the postorder traversal of its nodes' values.

For example:
Given binary tree {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

巴特西

LEETCODE —— Binary Tree的3 题 —— 3种非Recursive遍历

Binary Tree Preorder Traversal

Binary Tree Inorder Traversal

Binary Tree Postorder Traversal

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