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 (Java)
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> result = new ArrayList<>();
if (root == null) {
return result;
}
Stack<TreeNode> stack = new Stack<>();
TreeNode current = root;
while (current != null || !stack.isEmpty()) {
while (current != null) {
result.add(current.val);
stack.push(current.right);
current = current.left;
}
current = stack.pop();
}
return result;
}
}
Solution 1 (Javascript)
/**
* 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 (Javascript)
/**
* 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).