Skip to main content

Binary Tree Preorder Traversal

Problem statement​

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

Follow up: Recursive solution is trivial, could you do it iteratively?

My solution​

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[]}
 */
var preorderTraversal = function(root) {
  const result = []
  
  dfs(root)
    
    function dfs(node) {
        if (!node) {
            return
        }
        
        result.push(node.val)
        if (node.left) {
            dfs(node.left)
        }
        if (node.right) {
            dfs(node.right)
        }
    }
    
    return result
};