[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;
}
};