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94. Binary Tree Inorder Traversa

94. Binary Tree Inorder Traversa

作者: GoDeep | 来源:发表于2018-04-27 16:16 被阅读0次

    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?

    还是维护一个p指针吧,不要直接用stack里面的值

    class Solution:
        def inorderTraversal(self, root):
            """
            :type root: TreeNode
            :rtype: List[int]
            """
            if not root: return []
            res = []
            st = []
            p = root
            while p or st:
                while p:
                    st.append(p)
                    p = p.left
                t = st.pop()
                res.append(t.val)
                p = t.right
            return res
    

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