[LintCode]Binary Tree Preorder Traversal
答案
/**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
* @param root: The root of binary tree.
* @return: Preorder in ArrayList which contains node values.
*/
public ArrayList<Integer> preorderTraversal(TreeNode root) {
// 2015-3-22 DFS
ArrayList<Integer> rst = new ArrayList<>();
if (root == null) {
return rst;
}
// divide
ArrayList<Integer> left = preorderTraversal(root.left);
ArrayList<Integer> right = preorderTraversal(root.right);
// conquer
rst.add(root.val);
rst.addAll(left);
rst.addAll(right);
return rst;
}
}
本文介绍了一种实现二叉树前序遍历的方法,使用递归方式完成节点值的收集,并返回一个包含所有节点值的ArrayList。该算法首先访问根节点,然后递归地进行左子树和右子树的遍历。
407





