94. Binary Tree Inorder Traversal

https://leetcode.com/problems/binary-tree-inorder-traversal/

Problem

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?

Solution

1. Recursive

/**
 * 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> values;
        inorderTraversal(root, values);
        return values;
    }
    void inorderTraversal(TreeNode* root, vector<int> &values) {
        if (root == nullptr) return;
        inorderTraversal(root->left, values);
        values.push_back(root->val);
        inorderTraversal(root->right, values);
    }
    
};

2. Iterative

/**
 * 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> values; stack<TreeNode*> nodes;
        TreeNode* current = root;
        while (!nodes.empty() || current != nullptr) {
            if (current != nullptr) {
                nodes.push(current);
                current = current->left;
            } else {
                values.push_back(nodes.top()->val);
                current = nodes.top()->right;
                nodes.pop();
            }
        }
        return values;
    }
};

• #tree

• #inorder