LeetCode114 Flatten Binary Tree to Linked List

Given a binary tree, flatten it to a linked list in-place. （Medium）

For example,
Given

         1
        / 
       2   5
      /    
     3   4   6

The flattened tree should look like:

   1
    
     2
      
       3
        
         4
          
           5
            
             6

 

分析：

将树的问题和链表插入问题结合。对于每个节点，寻找左子树的最右端，它应该是左子树前序遍历的最后一位，也就是生成链表中，右子树根节点的前一位。

按照链表插入的方法处理即可。

代码：

/**
 * 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:
    void flatten(TreeNode* root) {
        if (root == nullptr) {
            return;
        }
        while (root != nullptr) {
            if (root -> left != nullptr) {
                TreeNode* temp = root -> left;
                while (temp -> right != nullptr) {
                    temp = temp -> right;
                }
                temp -> right = root -> right;
                root -> right = root -> left;
                root -> left = nullptr;
            }
            root = root -> right;
        }
    }
};