LeetCode----Invert Binary Tree

Invert Binary Tree

Invert a binary tree.

     4
   /   \
  2     7
 / \   / \
1   3 6   9
to
     4
   /   \
  7     2
 / \   / \
9   6 3   1
Trivia:
This problem was inspired by  this original tweet  by  Max Howell :
Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so fuck off.


分析:

这题是谷歌面试写出OS X中写出Homebrew软件的程序员的问题,结果他没有写出来,然后就悲剧了。我用递归和非递归都实现了一遍,然而并不代表我能去谷歌。

递归中,只需要递归地将当前节点的左右子节点交换即可。

非递归中,需要使用栈保存当前节点左右非空子节点,然后在循环中弹栈访问,并交换其左右子节点。


递归代码:

class Solution(object):
    def invertTree(self, root):
        """
        :type root: TreeNode
        :rtype: TreeNode
        """
        if not root:
            return None
        r = root.right
        l = root.left
        root.right = self.invertTree(l)
        root.left = self.invertTree(r)
        return root


非递归代码:

class Solution(object):
    def invertTree(self, root):
        """
        :type root: TreeNode
        :rtype: TreeNode
        """
        if not root:
            return None
        st = []
        p = root
        st.append(p)
        while st:
            p = st.pop()
            l = p.left
            r = p.right
            if l or r:
                p.left = r
                p.right = l
            if l:
                st.append(l)
            if r:
                st.append(r)
        return root


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