C++實現LeetCode(156.二叉樹的上下顛倒)

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[LeetCode] 156. Binary Tree Upside Down 二叉樹的上下顛倒

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.

Example:

Input: [1,2,3,4,5]

    1
/ \
2   3
/ \
4   5

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

   4
/ \
5   2
/ \
3   1  

Clarification:

Confused what [4,5,2,#,#,3,1] means? Read more below on how binary tree is serialized on OJ.

The serialization of a binary tree follows a level order traversal, where ‘#’ signifies a path terminator where no node exists below.

Here’s an example:

   1
/ \
2   3
/
4
\
5

The above binary tree is serialized as [1,2,3,#,#,4,#,#,5].

這道題讓我們把一棵二叉樹上下顛倒一下，而且限制瞭右節點要麼為空要麼一定會有對應的左節點。上下顛倒後原來二叉樹的最左子節點變成瞭根節點，其對應的右節點變成瞭其左子節點，其父節點變成瞭其右子節點，相當於順時針旋轉瞭一下。對於一般樹的題都會有迭代和遞歸兩種解法，這道題也不例外，先來看看遞歸的解法。對於一個根節點來說，目標是將其左子節點變為根節點，右子節點變為左子節點，原根節點變為右子節點，首先判斷這個根節點是否存在，且其有沒有左子節點，如果不滿足這兩個條件的話，直接返回即可，不需要翻轉操作。那麼不停的對左子節點調用遞歸函數，直到到達最左子節點開始翻轉，翻轉好最左子節點後，開始回到上一個左子節點繼續翻轉即可，直至翻轉完整棵樹，參見代碼如下：

解法一：

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;
    }
};

下面我們來看迭代的方法，和遞歸方法相反的時，這個是從上往下開始翻轉，直至翻轉到最左子節點，參見代碼如下：

解法二：

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;
    }
};

Github 同步地址：

https://github.com/grandyang/leetcode/issues/156

類似題目：

Reverse Linked List

參考資料：

https://leetcode.com/problems/binary-tree-upside-down/

https://leetcode.com/problems/binary-tree-upside-down/discuss/49412/Clean-Java-solution

https://leetcode.com/problems/binary-tree-upside-down/discuss/49432/Easy-O(n)-iteration-solution-Java

https://leetcode.com/problems/binary-tree-upside-down/discuss/49406/Java-recursive-(O(logn)-space)-and-iterative-solutions-(O(1)-space)-with-explanation-and-figure

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