Day 43 LeetCode 94 Binary Tree Inorder Traversal.cpp

/*
94. Binary Tree Inorder Traversal
Solved
Easy

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 = [1,2,3,4,5,null,8,null,null,6,7,9]
Output: [4,2,6,5,7,1,3,9,8]


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

Example 4:
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?
*/

/**
 * 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:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> res;
        inorder(root, res);
        return res;        
    }

  private:
    void inorder(TreeNode* node, vector<int>& res) {
        if (!node) {
            return;
        }
        inorder(node->left, res);
        res.push_back(node->val);
        inorder(node->right, res);
    }    
};