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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 pre-order traversal of the original tree.


Let's consider an example to understand the problem. Given the following binary tree:

        /  \
       /    \
      6      8
     / \    / \
    2   3  7   5
   /         \   \
  9           -6  13

After extracting the leaf nodes, the pre-order 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 pre-order 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.


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 the BinaryTree class to keep track of the front of the doubly linked list.
  • The extractLeaves function takes a node 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 non-empty 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 pre-order 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

Time Complexity

  • The algorithm involves a pre-order traversal of the binary tree.
  • During the traversal, each node is processed once, and constant-time 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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