Question
Given the root of a binary tree, return the preorder traversal of its nodes’ values.
Example 1:
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| Input: root = [1,null,2,3]
Output: [1,2,3]
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Example 2:
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| Input: root = []
Output: []
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Example 3:
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| Input: root = [1]
Output: [1]
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Solution
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| class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<>();
Stack<TreeNode> stack = new Stack<>();
if(root != null) stack.push(root);
while(!stack.isEmpty()){
root = stack.pop();
res.add(root.val);
if(root.right != null) stack.push(root.right);
if(root.left != null) stack.push(root.left);
}
return res;
}
}
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Complexity:
Time complexity: O(n)
Space complexity: O(n)
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| class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<>();
helper(root, res);
return res;
}
private void helper(TreeNode root, List<Integer> res) {
if (root == null) return;
res.add(root.val);
helper(root.left, res);
helper(root.right, res);
}
}
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Complexity:
Time complexity: O(n)
Space complexity: O(n)