Binary Tree Inorder Traversal LeetCode Solution | Easy Approach

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?

 Binary Tree Inorder Traversal Solutions

✅Time: O(n)
✅Space: O(n)

C++

class Solution {
 public:
  vector<int> inorderTraversal(TreeNode* root) {
    vector<int> ans;
    stack<TreeNode*> stack;

    while (root || !stack.empty()) {
      while (root) {
        stack.push(root);
        root = root->left;
      }
      root = stack.top(), stack.pop();
      ans.push_back(root->val);
      root = root->right;
    }

    return ans;
  }
};

Java

 class Solution {
  public List<Integer> inorderTraversal(TreeNode root) {
    List<Integer> ans = new ArrayList<>();
    Deque<TreeNode> stack = new ArrayDeque<>();

    while (root != null || !stack.isEmpty()) {
      while (root != null) {
        stack.push(root);
        root = root.left;
      }
      root = stack.pop();
      ans.add(root.val);
      root = root.right;
    }

    return ans;
  }
}

Python

class Solution:
  def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
    ans = []
    stack = []

    while root or stack:
      while root:
        stack.append(root)
        root = root.left
      root = stack.pop()
      ans.append(root.val)
      root = root.right

    return ans

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