156. Binary Tree Upside Down (Medium)

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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

Solution 1: recursive 3ms

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：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
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;
    }
};

Solution 2: 3ms

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：

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