[LeetCode] 114. Flatten Binary Tree to Linked List

Given a binary tree, flatten it to a linked list in-place.
For example,
Given
         1
        / \
       2   5
      / \   \
     3   4   6
The flattened tree should look like:
   1
    \
     2
      \
       3
        \
         4
          \
           5
            \
             6
click to show hints.
Hints:
If you notice carefully in the flattened tree, each node's right child points to the next node of a pre-order traversal.
Thought process:
Each node's right sub-tree is linked to its largest left child. So the idea is to recursively flatten a node's right sub-tree, then flatten its left sub-tree, and finally link the right sub-tree after the left sub-tree and move the sub-tree to the right.
The key point is to keep a prev pointer that points to the previously finished node, and attach it to the current node.

Solution 1 (C++):
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/**
 * 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 {
private:
    TreeNode* prev = NULL;
    
public:
    void flatten(TreeNode* root) {
        if (root == NULL) {
            return;
        }
        
        flatten(root->right);
        flatten(root->left);
        
        root->right = prev;
        root->left = NULL;
        prev = root;
    }
};

Solution 2 (Java):

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private TreeNode previous = null;
    
    public void flatten(TreeNode root) {
        if (root == null) {
            return;
        }
        
        flatten(root.right);
        flatten(root.left);
        
        root.right = previous;
        root.left = null;
        previous = root;
    }
}

Time complexity: O(n).
Location: Philadelphia, PA, USA

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