[leetcode] 156. Binary Tree Upside Down 解题报告

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

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  

思路: 题意是说每个结点的右子树要么为空, 要么一定有一个左子树孩子和一个右子树孩子, 因此树的形状是左偏的. 所以我们要将最左边的子树作为最终的新根结点, 然后递归的将其父结点作为其右孩子,并且父结点的右孩子作为其左孩子. 一个非常重要的地方是每次一定要将父结点的左右孩子都置为空, 因为父结点设置成其左孩子的右孩子之后成了叶子结点, 需要将其指针断掉.

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==NULL || root->left==NULL) return root;
        auto left = upsideDownBinaryTree(root->left);
        root->left->right = root;
        root->left->left = root->right;
        root->left = NULL, root->right = NULL;
        return left;
    }
};