114. Flatten Binary Tree to Linked List

Problem Statement

Given the root of a binary tree, flatten the tree into a "linked list":

• The "linked list" should use the same TreeNode class where the right child pointer points to the next node in the list and the left child pointer is always null.

• The "linked list" should be in the same order as a pre-order traversal of the binary tree.

Example 1:

Input: root = [1,2,5,3,4,null,6]
Output: [1,null,2,null,3,null,4,null,5,null,6]

Example 2:

Input: root = []
Output: []

Example 3:

Input: root = [0]
Output: [0]

Constraints:

• The number of nodes in the tree is in the range [0, 2000].

• -100 <= Node.val <= 100

Follow up: Can you flatten the tree in-place (with O(1) extra space)?

Intuition

Approach 1:
Using recursion, maintain a prev node to attach left tree to right part
And right part to null

Approach 2:
Iterative, same as above

Approach 3: 
Morris traversal

Links

https://leetcode.com/problems/flatten-binary-tree-to-linked-list/description/

Video Links

https://www.youtube.com/watch?v=sWf7k1x9XR4&t=587s&ab_channel=takeUforward

Approach 1:

Recursion

C++

class Solution {
public:
    TreeNode *prev = nullptr;
    void flatten(TreeNode* root) {
        if(root == nullptr)
            return ;

        flatten(root->right);
        flatten(root->left);

        root->right = prev;
        root->left = nullptr;
        prev = root;
    }
};

Approach 2:

Iterative

C++

class Solution {
public:
    
    void flatten(TreeNode* root) {
        if(root == NULL) return;
        stack<TreeNode*> st;
        st.push(root);

        while(!st.empty()){
            TreeNode* curr = st.top();
            st.pop();

            if(curr->right)
                st.push(curr->right);
            if(curr->left)
                st.push(curr->left);

            if(!st.empty())
                curr->right = st.top();

            curr->left = nullptr;
        }
    }
};

Approach 3:

C++

Approach 4:

C++

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