LeetCode-in-All

94. Binary Tree Inorder Traversal

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 = []

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:

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

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:
    vector<int> a;
    void inord(TreeNode* root)
    {
        if(!root) return;
        if(root->left) inord(root->left);
        a.push_back(root->val);
        if(root->right) inord(root->right);
    }
    vector<int> inorderTraversal(TreeNode* root) {
        inord(root);
        return a;
    }
};