Problem
Given a binary tree, return the preorder traversal of its nodes' values.
Example:
Input: [1,null,2,3]
1
\
2
/
3
Output: [1,2,3]
Follow up: Recursive solution is trivial, could you do it iteratively?
Solution 1
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @return {number[]}
*/
var preorderTraversal = function(root) {
var res = [];
helper(root, res);
return res;
};
var helper = function (root, res) {
if (!root) return;
res.push(root.val);
helper(root.left, res);
helper(root.right, res);
};
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).
Solution 2
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @return {number[]}
*/
var preorderTraversal = function(root) {
if (!root) return [];
var res = [];
var queue = [root];
var node = null;
while (queue.length) {
node = queue.pop();
res.push(node.val);
if (node.right) queue.push(node.right);
if (node.left) queue.push(node.left);
}
return res;
};
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).