From 87dfb6a8b7ef47b66bf25aee9c39a7d4760bb40b Mon Sep 17 00:00:00 2001
From: Anshika Yadav <14anshika7yadav@gmail.com>
Date: Tue, 11 Jun 2024 13:05:13 +0530
Subject: [PATCH] Added  Leetcode Problem : 114 #957

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+---
+id: flatten-binary-tree-to-linked-list
+title: Flatten Binary Tree To Linked List
+sidebar_label: 0114 Flatten Binary Tree To Linked List
+tags:
+  - tree
+  - tree traversal
+  - Linked List
+  - LeetCode
+  - C++
+description: "This is a solution to the Flatten Binary Tree To Linked List problem on LeetCode."
+---
+
+## Problem Description
+
+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.
+
+### Examples
+
+**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 2:**
+
+```
+Input: root = [0]
+Output: [0]
+```
+
+### Constraints
+
+- The number of nodes in the tree is in the range $[0, 2000]$.
+- $-100 \leq \text{Node.val} \leq 100$.
+
+### Approach 
+
+To solve this problem(flatten binary tree to linked list) first we will store binary tree nodes value in a vector using preorder traversal then we will convert this vector into another tree (simply linked list) which has no left child only right child as given in question description or we can also use recursion as used in java solution given below.
+
+#### Code in C++
+
+```cpp
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
+ * right(right) {}
+ * };
+ */
+class Solution {
+public:
+    
+    //Funtion to store tree node value in vector using preorder traversal
+    void preorder(TreeNode* root, vector<int>& a) {
+        if (root == NULL) {
+            return;
+        }
+        a.push_back(root->val);
+        preorder(root->left, a);
+        preorder(root->right, a);
+    }
+
+    void flatten(TreeNode* root) { // to flatten binary tree to linked list
+        if (root == NULL) {
+            return;
+        }
+        vector<int> a;
+        preorder(root, a);
+        TreeNode* temp = root;
+        root->left = NULL;
+        int i = 1;
+        while (i < a.size()) {
+            temp->right = new TreeNode(a[i++]);
+            temp = temp->right;
+        }
+    }
+};
+```
+
+
+#### Code in Java
+
+```java
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public void flatten(TreeNode root) {
+        if(root==null ||root.left==null && root.right==null)
+            return;
+        else{
+            if(root.left!=null){
+                TreeNode temp=root.right;
+                root.right=root.left;
+                root.left=null;
+                TreeNode temp1=root.right;
+                while(temp1.right!=null)
+                    temp1=temp1.right;
+                temp1.right=temp;
+            }
+            flatten(root.right);
+        }
+    }
+}
+
+```
+