Binary Tree Inorder Traversal
Problem Statement - link #
Given the root of a binary tree, return the inorder traversal of its nodes’ values.
Examples #
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
Solutions #
/**
* 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 {
vector<int> ans;
public:
void helper(TreeNode* root)
{
if(!root) return;
helper(root->left);
ans.push_back(root->val);
helper(root->right);
}
vector<int> inorderTraversal(TreeNode* root) {
helper(root);
return ans;
}
};