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94. Binary Tree Inorder Traversal

Posted on Edited on In Disqus:
Symbols count in article: 1.7k Reading time ≈ 2 mins.

iterative stack O(n) time O(n) space

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        stack<pair<TreeNode *, bool>> s;
        s.emplace(root, false);
        vector<int> res;
        while (!s.empty()) {
            auto [n, v] = s.top();
            s.pop();
            if (!n) continue;
            if (v) {
                res.push_back(n->val);
            } else {
                s.emplace(n->right, false);
                s.emplace(n, true);
                s.emplace(n->left, false);
            }
        }
        return res;
    }
};

recursive O(n) time

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> res;
        dfs(root, res);
        return res;
    }

    void dfs(TreeNode *root, vector<int> &res) {
        if (!root) return;
        dfs(root->left, res);
        res.push_back(root->val);
        dfs(root->right, res);
    }
};
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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> res;
        if (!root) return {};
        for (int x : inorderTraversal(root->left)) {
            res.push_back(x);
        }
        res.push_back(root->val);
        for (int x : inorderTraversal(root->right)) {
            res.push_back(x);
        }
        return res;
    }
};