[LeetCode] Binary Tree Upside Down 二叉树的上下颠倒

 

Given a binary tree where all the right nodes are either leaf nodes with a sibling (a left node that shares the same parent node) or empty, flip it upside down and turn it into a tree where the original right nodes turned into left leaf nodes. Return the new root.

For example:

Given a binary tree {1,2,3,4,5},

    1

   / \

  2   3

 / \

4   5

return the root of the binary tree [4,5,2,#,#,3,1].

   4

  / \

 5   2

    / \

   3   1  

 

这道题让我们把一棵二叉树上下颠倒一下,而且限制了右节点要么为空要么一定会有对应的左节点。上下颠倒后原来二叉树的最左子节点变成了根节点,其对应的右节点变成了其左子节点,其父节点变成了其右子节点,相当于顺时针旋转了一下。对于一般树的题都会有迭代和递归两种解法,这道题也不例外,那么我们先来看看递归的解法。对于一个根节点来说,我们的目标是将其左子节点变为根节点,右子节点变为左子节点,原根节点变为右子节点,那么我们首先判断这个根节点是否存在,且其有没有左子节点,如果不满足这两个条件的话,直接返回即可,不需要翻转操作。那么我们不停的对左子节点调用递归函数,直到到达最左子节点开始翻转,翻转好最左子节点后,开始回到上一个左子节点继续翻转即可,直至翻转完整棵树,参见代码如下:

 

解法一:

// Recursion
class Solution {
public:
    TreeNode *upsideDownBinaryTree(TreeNode *root) {
        if (!root || !root->left) return root;
        TreeNode *l = root->left, *r = root->right;
        TreeNode *res = upsideDownBinaryTree(l);
        l->left = r;
        l->right = root;
        root->left = NULL;
        root->right = NULL;
        return res;
    }
};

 

下面我们来看迭代的方法,和递归方法相反的时,这个是从上往下开始翻转,直至翻转到最左子节点,参见代码如下:

 

解法二:

// Iterative
class Solution {
public:
    TreeNode *upsideDownBinaryTree(TreeNode *root) {
        TreeNode *cur = root, *pre = NULL, *next = NULL, *tmp = NULL;
        while (cur) {
            next = cur->left;
            cur->left = tmp;
            tmp = cur->right;
            cur->right = pre;
            pre = cur;
            cur = next;
        }
        return pre;
    }
};

 

参考资料:

 

https://discuss.leetcode.com/topic/5961/easy-o-n-iteration-solution-java

https://discuss.leetcode.com/topic/40924/java-recursive-o-logn-space-and-iterative-solutions-o-1-space-with-explanation-and-figure

 

    原文作者:Grandyang
    原文地址: http://www.cnblogs.com/grandyang/p/5172838.html
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
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