Leetcode解题-Binary Tree Inorder Traversal

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描述

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?

分析

中序遍历，基本上沿用上一题的框架。

Morris遍历继续参考这篇文章。

代码

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class Solution(object):
    def inorderTraversalR(self, root, rs):
        if not root:
            return
        self.inorderTraversalR(root.left, rs)
        rs.append(root.val)
        self.inorderTraversalR(root.right, rs)

    def inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        rs = []
        self.inorderTraversalR(root, rs)
        return rs

迭代

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class Solution(object):
    def inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        rs = []
        ss = []
        cur = root
        while cur or ss:
            if cur:
                ss.append(cur)
                cur = cur.left
            else:
                cur = ss.pop()
                rs.append(cur.val)
                cur = cur.right
        return rs

Morris

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class Solution(object):
    def inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        rs = []
        cur = root
        while cur:
            if not cur.left:
                rs.append(cur.val)
                cur = cur.right
            else:
                # look up predecessor in left sub-tree
                node = cur.left
                while node.right and node.right != cur:
                    node = node.right
                if not node.right:
                    # make threaded
                    node.right = cur
                    cur = cur.left
                else:
                    # already threaded
                    node.right = None
                    rs.append(cur.val)
                    cur = cur.right
        return rs

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