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0094 - Binary Tree Inorder Traversal (Easy)

Problem Statement​

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?

Approach 1: Inorder Traversal​

Written by @wingkwong
/**
 * 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) {}
 * };
 */

// Time Complexity: O(N)
// Space Complexity: O(N)

// This is a standard in-order traversal problem, I'd suggest to learn pre-order and post-order as well.
// Here's a short tutorial if you're interested.
// https://wingkwong.github.io/leetcode-the-hard-way/tutorials/graph-theory/binary-tree
// then you may try the following problems 
// 144. Binary Tree Preorder Traversal: https://leetcode.com/problems/binary-tree-preorder-traversal/
// 145. Binary Tree Postorder Traversal: https://leetcode.com/problems/binary-tree-postorder-traversal/

class Solution {
public:
    vector<int> ans;
    void inorder(TreeNode* node) {
        if (node == NULL) return;
        // traverse the left node
        inorder(node->left);
        // do something with node value here
        ans.push_back(node->val);
        // traverse the right node
        inorder(node->right);
    }
    
    vector<int> inorderTraversal(TreeNode* root) {
        inorder(root);
        return ans;
    }
};

Approach 2: Iterative​

Written by @vigneshshiv
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
// Time complexity: O(n), where n - # of nodes in the tree
// Space complexity: O(n)
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> result = new LinkedList<>();
        Stack<TreeNode> stack = new Stack<>();
        while (root != null || !stack.isEmpty()) {
            // Keep traversing to left and add it to stack till last
            if (root != null) {
                stack.push(root);
                root = root.left;
            } else {
                //   1
                //  / \
                // 2   3
                // All left tree traversed and currently root is NULL 
                // 2 is added to result, Stack is only having 1
                // If current node is 2, and it's popped out, 2's right is assigned to root which is NULL
                // So in the next iteration root still be NULL. 
                // Stack top value 1 added to result and 1 popped out and 1's right 3 assigned to root. 
                result.add(stack.peek().val);
                root = stack.pop().right;
            }
        }
        return result;
    }
}