Binary Tree Inorder Traversal LeetCode Solution

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

Binary Tree Inorder Traversal LeetCode Solution in Java

public List<Integer> inorderTraversal(TreeNode root) {
    List<Integer> list = new ArrayList<Integer>();

    Stack<TreeNode> stack = new Stack<TreeNode>();
    TreeNode cur = root;

    while(cur!=null || !stack.empty()){
        while(cur!=null){
            stack.add(cur);
            cur = cur.left;
        }
        cur = stack.pop();
        list.add(cur.val);
        cur = cur.right;
    }

    return list;
}

Binary Tree Inorder Traversal LeetCode Solution in Python

def preorder(root):
  return [root.val] + preorder(root.left) + preorder(root.right) if root else []

Binary Tree Inorder Traversal LeetCode Solution in C++

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

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