easy LNR concept..see once

Given the root of a binary tree, return the inorder traversal of its nodes' values.

 

Example 1:


Input: root = [1,null,2,3]
Output: [1,3,2]
Example 2:

Input: root = []
Output: []
Example 3:

Input: root = [1]
Output: [1]
 

Constraints:

The number of nodes in the tree is in the range [0, 100].
-100 <= Node.val <= 100
 

Follow up: Recursive solution is trivial, could you do it iteratively?

Author: Gauranshgoel123

binary-tree-inorder-traversal.cpp

@@ -0,0 +1,29 @@
+/**
+ * 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:
+
+    void inOrder(TreeNode* root,vector<int> &ans){
+        if(root==nullptr){
+            return;
+        }
+        inOrder(root->left,ans);
+        ans.push_back(root->val);
+        inOrder(root->right,ans);
+    }
+    vector<int> inorderTraversal(TreeNode* root) {
+        vector<int> ans;
+        inOrder(root,ans);
+        return ans;
+    }
+};
+