94. Binary Tree Inorder Traversal Leetcode Python

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.

这道题目主要考察inorder traverse, 题目很简单，但是也要考虑traverse 和iterative 的方法。其中iterative 的方法参考了http://jelices.blogspot.com/2014/06/leetcode-python-binary-tree-inorder.html

首先是traversal

# 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 inorder(self,root,solution):
        if root:
            self.inorder(root.left,solution)
            solution.append(root.val)
            self.inorder(root.right,solution)
    def inorderTraversal(self, root):
        solution=[]
        self.inorder(root,solution)
        return solution
        

iterative 的方法

# 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=[]
        solution=[]
        node=root
        while node or len(stack)>0:
            if node!=None:
                stack.append(node)
                node=node.left
            else:
                node=stack.pop()
                solution.append(node.val)
                node=node.right
        return solution