Extract leaves of a binary tree in a doubly linked list
The problem discussed in this article involves extracting the leaves of a binary tree and arranging them in a doubly linked list. A binary tree is a hierarchical data structure where each node has at most two children – a left child and a right child. The problem requires extracting the leaf nodes (nodes with no children) from the binary tree and transforming them into a doubly linked list.
Problem Statement
Given a binary tree, we need to extract the leaf nodes from it and arrange them in a doubly linked list. The order of nodes in the linked list should correspond to the preorder traversal of the original tree.
Example
Let's consider an example to understand the problem. Given the following binary tree:
10
/ \
/ \
6 8
/ \ / \
2 3 7 5
/ \ \
9 6 13
After extracting the leaf nodes, the preorder traversal of the transformed tree will be: 10, 6, 2, 8, 7, 5. The extracted leaves will be arranged in a doubly linked list: 9, 3, 6, 13.
Idea to Solve
To solve this problem, we need to perform a preorder traversal of the binary tree while extracting the leaf nodes. During the traversal, when we encounter a leaf node, we remove it from the tree and transform it into a doubly linked list node. We also adjust the links of the nodes accordingly.
Pseudocode
Here's the pseudocode that outlines the algorithm to extract leaves of a binary tree in a doubly linked list:
function extractLeaves(node):
if node is not null:
if node.left is null and node.right is null:
node.right = front
if front is not null:
front.left = node
front = node
return null
node.right = extractLeaves(node.right)
node.left = extractLeaves(node.left)
return node
Algorithm Explanation
 We start by defining a
front
pointer in theBinaryTree
class to keep track of the front of the doubly linked list.  The
extractLeaves
function takes anode
as input and operates recursively.  When a leaf node is encountered (both left and right children are null), we transform it into a doubly linked list node.
 We adjust the right and left pointers of the node to connect it with the previous and next nodes in the linked list.
 If the
front
pointer is not null (indicating a nonempty list), we also adjust the left pointer of the front node.  The function then returns null, indicating that the leaf node should be removed from the tree.
 During the recursion, we traverse the tree in a preorder fashion.
 At the end of the algorithm, the binary tree will be modified, and the linked list of leaf nodes will be created.
Code Solution

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Time Complexity
 The algorithm involves a preorder traversal of the binary tree.
 During the traversal, each node is processed once, and constanttime operations are performed to adjust the links.
 Therefore, the time complexity of the algorithm is O(n), where n is the number of nodes in the binary tree.
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