Problem description:

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

For example, given the following tree:

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2
3
4
5
    1
   / \
  2   5
 / \   \
3   4   6

The flattened tree should look like:

1
2
3
4
5
6
7
8
9
10
11
1
 \
  2
   \
    3
     \
      4
       \
        5
         \
          6

Solution:

Recursive solution:

Use a global variable to store a previous node, and traverse the tree with post-order traversal.

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prev= NULL

  1
 / \
2   3


prev= 3

  1
 / 
2 
 \
  3

prev= 2
  1
   \
    2
     \
      3

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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 {
public:
    TreeNode* prev= NULL;
    
    void flatten(TreeNode* root) {
        if(!root) return;
        if(root->right) flatten(root->right);
        if(root->left) flatten(root->left);
        
        root->right= prev;
        root->left= NULL;
        prev= root;
    }
};

time complexity: $O()$
space complexity: $O()$
reference: