Given a binary tree, return the preorder traversal of its nodes' values.
For example:
Given binary tree {1,#,2,3}
,
1 \ 2 / 3
return [1,2,3]
.
Note: Recursive solution is trivial, could you do it iteratively? 递归是微不足道的,你能使用迭代完成吗?
方法一 递归
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> result = new ArrayList<>();
preorderTraversal(root, result);
return result;
}
public void preorderTraversal(TreeNode root, List<Integer> list) {
if (root != null) list.add(root.val);
else return;
if (root.left != null) preorderTraversal(root.left, list);
if (root.right != null) preorderTraversal(root.right, list);
}
方法二 使用栈,非递归
public List<Integer> preorderTraversal2(TreeNode root) {
List<Integer> list = new ArrayList<>();
if (root == null) return list;
Stack<TreeNode> stack = new Stack<>();
stack.push(root);
while (!stack.isEmpty()) {
TreeNode tmp = stack.pop();
list.add(tmp.val);
if (tmp.right != null) stack.push(tmp.right);
if (tmp.left != null) stack.push(tmp.left);
}
return list;
}