LeetCode #156. Binary Tree Upside Down

最新推荐文章于 2021-08-08 02:32:16 发布

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本文详细解析了一种特殊的二叉树翻转算法，即倒置二叉树，将原树的右节点转换为左叶节点，通过迭代或递归方式实现。文章通过实例展示了算法的具体操作过程，并解释了二叉树的序列化方式。

题目描述：

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:

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

class Solution {
public:
    TreeNode* upsideDownBinaryTree(TreeNode* root) {
        // 右节点要么为空，要么为叶节点，此时左节点一定不为空
        if(root==NULL) return NULL;
        else if(root->left==NULL&&root->right==NULL) return root;
        // 用迭代的方法需要维护当前节点，上一个节点，上一个节点的右节点，下一个节点
        // 每次迭代当前节点的右节点是上一个节点，左节点是上一个节点的右节点
        TreeNode* pre=NULL, *cur=root, *next=root->left, *pre_right=NULL;
        while(cur)
        {
            next=cur->left;
            cur->left=pre_right;
            pre_right=cur->right;
            cur->right=pre;
            pre=cur;
            cur=next;
        }
        return pre; // cur为空时，pre就是最后访问的节点
    }
};

class Solution {
public:
    TreeNode* upsideDownBinaryTree(TreeNode* root) {
        // 右节点要么为空，要么为叶节点，此时左节点一定不为空
        if(root==NULL) return NULL;
        else if(root->left==NULL&&root->right==NULL) return root;
        TreeNode* left=root->left;
        TreeNode* right=root->right;
        TreeNode* result=upsideDownBinaryTree(left);
        left->left=right;
        left->right=root;
        root->left=NULL;
        root->right=NULL;
        return result;
    }
};