114. Flatten Binary Tree to Linked List

问题：

拉平二叉树 -> 只有right分支的链表。

For example, given the following tree:
    1
   / 
  2   5
 /    
3   4   6
The flattened tree should look like:
1
 
  2
   
    3
     
      4
       
        5
         
          6

　　

解法：Binary Tree（二叉树）

递归做法：

• base：root==null，直接返回
• 对每个节点node：
  • 递归拉平node->left，node->right
  • 对已经拉平的两棵子树：right，left
    • node->right指向原来的left，node->left=null
    • node->right的最后一个right节点->right指向原来的right。

代码参考：

/**
 * 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:
    //right = node.right; left = node.left;
    //1.node.right = left
    //2.node.right.last.right = right
    void flatten(TreeNode* root) {
        if(!root) return;
        
        flatten(root->left);
        flatten(root->right);
        
        TreeNode* left = root->left;
        TreeNode* right = root->right;
        
        //1.node.right = left
        root->left = nullptr;
        root->right = left;
        //2.node.right.last.right = right
        TreeNode* p = root;
        while(p->right) {
            p = p->right;
        }
        p->right = right;
    }
};