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094--Binary Tree Inorder Traversal.md

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

Example:

Input: [1,null,2,3]
   1
    \
     2
    /
   3

Output: [1,3,2]
Follow up: Recursive solution is trivial, could you do it iteratively?
# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        
        res = []
        self.helper(root, res)
        return res
    def helper(self, root, res):
        if root != None:
            if root.left != None:
                self.helper(root.left, res)
            res.append(root.val)
            if root.right != None:
                self.helper(root.right, res)
# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        
        res = []
        stack = []
        curr = root
        while curr!=None or stack!=[]:
            while curr!=None:
                stack.append(curr)
                curr = curr.left
            curr = stack.pop()
            res.append(curr.val)
            curr = curr.right
        return res