leetcode--114--二叉树展开为链表

题目：
给定一个二叉树，原地将它展开为一个单链表。

例如，给定二叉树

1

/
2 5
/ \
3 4 6
将其展开为：

1

2

3

4

5

6

链接：https://leetcode-cn.com/problems/flatten-binary-tree-to-linked-list

思路：
1、采用后序遍历的思路，左右根。
2、具体执行流程见代码注释

Python代码：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def flatten(self, root):
        """
        :type root: TreeNode
        :rtype: None Do not return anything, modify root in-place instead.
        """
        if not root:
            return root
        if (not root.left and not root.right):
            return root
        
        self.flatten(root.left)  # 展开左右子树
        self.flatten(root.right)

        temp = root.right  # 备份右子树
        root.right = root.left  # 将root的右指针指向展开的左子树
        root.left = None  # 将root的左子树置空
        while root.right:
            root = root.right
        root.right = temp # 将展开过的右子树拼接在root当前的叶子节点后面

C++代码：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    void flatten(TreeNode* root) {
        if(root==nullptr){
            return;
        }
        if(root->left==NULL && root->right==NULL){
            return;
        }

        flatten(root->left);
        flatten(root->right);
        auto temp = root->right;
        root->right = root->left;
        root->left = NULL;
        while (root->right!=NULL){
            root = root->right;
        }
        root->right = temp;
    }
};