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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?
难度：medium
题目：给定二叉树，返回其前序遍历结点值。（不要使用递归）
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) {val = x;}
* }
*/
class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
Stack<TreeNode> stack = new Stack<>();
List<Integer> result = new ArrayList<>();
// root, left, right
while (root != null || !stack.isEmpty()) {
if (null == root) {
root = stack.pop();
}
result.add(root.val);
if (root.right != null) {
stack.push(root.right);
}
root = root.left;
}
return result;
}
}
