Binary Tree Inorder Traversal

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]

Example 4:

Input: root = [1,2]
Output: [2,1]

Example 5:

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

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?

Solutions

🧠 Cpp

/**
 * Definition for a binary tree node.
 * struct TreeNode
 {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution
{
public:
    vector<int> inorderTraversal(TreeNode* root)
    {
        vector<int> res;

        if(!root)
            return res;
        if(root->left)
        {
            vector<int> &&tres = inorderTraversal(root->left);
            res.insert(res.end(), tres.begin(), tres.end());
        }

        res.emplace_back(root->val);

        if(root->right)
        {
            vector<int> &&tres = inorderTraversal(root->right);
            res.insert(res.end(), tres.begin(), tres.end());
        }

        return res;
    }
};