原题链接
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?
思路解析
- 非递归遍历通常都是使用栈来完成，（递归的本质也是栈）前序遍历二叉树，先访问节点，然后把右子树入栈，再把左子树入栈， 如此循环即可
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> list = new ArrayList<>();
if (root == null) {
return list;
}
Stack<TreeNode> stack = new Stack<>();
stack.push(root);
while(!stack.isEmpty()) {
root = stack.pop();
list.add(root.val);
if (root.right != null) {
stack.push(root.right);
}
if (root.left != null) {
stack.push(root.left);
}
}
return list;
}
