94. Binary Tree Inorder Traversal

94. Binary Tree Inorder Traversal

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

For example:
Given binary tree [1,null,2,3],

   1
    
     2
    /
   3

return [1,3,2].

该题是做树的中序遍历，下面分别是递归解法和非递归解法：

递归解法：

class Solution(object):
    def inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if not root:
            return []
        return self.inorderTraversal(root.left) + [root.val] + self.inorderTraversal(root.right)

 非递归解法：

class Solution(object):
    def inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if not root:
            return []
        stack = []
        node = root
        result = []　　　
        while node or len(stack) > 0:
            while node:
                stack.append(node)
                node = node.left
            node = stack.pop()
            result.append(node.val)
            node = node.right
        return result