94. Binary Tree Inorder Traversal

Recursion

/*
 * @lc app=leetcode id=94 lang=cpp
 *
 * [94] Binary Tree Inorder Traversal
 *
 * https://leetcode.com/problems/binary-tree-inorder-traversal/description/
 *
 * algorithms
 * Easy (77.38%)
 * Likes:    13994
 * Dislikes: 833
 * Total Accepted:    3M
 * Total Submissions: 3.9M
 * Testcase Example:  '[1,null,2,3]'
 *
 * 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]
 *
 * Explanation:
 *
 *
 *
 *
 * 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]
 *
 * Explanation:
 *
 *
 *
 *
 * 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?
 */

// @lc code=start
/**
 * 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 {
private:
    vector<int> res;
public:
    vector<int> inorderTraversal(TreeNode* root) {
        dfs(root);
        return res;
    }
    void dfs(TreeNode* node) {
        if (!node) return;
        dfs(node->left);
        res.emplace_back(node->val);
        dfs(node->right);
    }
};
// @lc code=end

• T: $O(N)$
• S: $O(N)$

Iteration

class Solution {
private:
    vector<int> res;
public:
    vector<int> inorderTraversal(TreeNode* root) {
        stack<TreeNode*> stk;
        vector<int> res;
        TreeNode* node = root;
        while (node || !stk.empty()) {
            while (node) {
                stk.push(node);
                node = node->left;
            }
            node = stk.top(); stk.pop();
            res.emplace_back(node->val);
            node = node->right;
        }
        return res;
    }
};

• T: $O(N)$
• S: $O(N)$