Binary Tree Preorder Traversal LeetCode Solution

Problem – Binary Tree Preorder Traversal LeetCode Solution

Given the root of a binary tree, return the preorder traversal of its nodes’ values.

Example 1:

Input: root = [1,null,2,3]
Output: [1,2,3]

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 Preorder Traversal LeetCode Solution in Java

public List<Integer> preorderTraversal(TreeNode node) {
	List<Integer> list = new LinkedList<Integer>();
	Stack<TreeNode> rights = new Stack<TreeNode>();
	while(node != null) {
		list.add(node.val);
		if (node.right != null) {
			rights.push(node.right);
		}
		node = node.left;
		if (node == null && !rights.isEmpty()) {
			node = rights.pop();
		}
	}
    return list;
}

Binary Tree Preorder Traversal LeetCode Solution in Python

def preorderTraversal(self, root):
    ret = []
    stack = [root]
    while stack:
        node = stack.pop()
        if node:
            ret.append(node.val)
            stack.append(node.right)
            stack.append(node.left)
    return ret

Binary Tree Preorder Traversal LeetCode Solution in C++

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

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