94. Binary Tree Inorder Traversal | Leetcode solution

 94. Binary Tree Inorder Traversal

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

solution:

/**

 * 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>&result)

    {   

        if(!root)

        {

            return ;

        }

        inorder(root->left , result);

        result.push_back(root->val);

        inorder(root->right , result);

        

    }

    vector<int> inorderTraversal(TreeNode* root) {

        vector<int>result;

        inorder(root , result);

        return result;

    }

};