94

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]

 

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?

Companies: Adobe, Amazon, Facebook

Related Topics:
Stack, Tree, Depth-First Search, Binary Tree

Similar Questions:

• Validate Binary Search Tree (Medium)
• Binary Tree Preorder Traversal (Easy)
• Binary Tree Postorder Traversal (Easy)
• Binary Search Tree Iterator (Medium)
• Kth Smallest Element in a BST (Medium)
• Closest Binary Search Tree Value II (Hard)
• Inorder Successor in BST (Medium)
• Convert Binary Search Tree to Sorted Doubly Linked List (Medium)
• Minimum Distance Between BST Nodes (Easy)

Solution 1. Recursive

// OJ: https://leetcode.com/problems/binary-tree-inorder-traversal/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(H)
class Solution {
private:
    vector<int> ans;
    void dfs(TreeNode* root) {
        if (!root) return;
        dfs(root->left);
        ans.push_back(root->val);
        dfs(root->right);
    }
public:
    vector<int> inorderTraversal(TreeNode* root) {
        dfs(root);
        return ans;
    }
};

Solution 2. Iterative

// OJ: https://leetcode.com/problems/binary-tree-inorder-traversal/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(H)
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> ans;
        stack<TreeNode*> s;
        while (root || s.size()) {
            while (root) {
                s.push(root);
                root = root->left;
            }
            root = s.top();
            s.pop();
            ans.push_back(root->val);
            root = root->right;
        }
        return ans;
    }
};