题目描述
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?
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
import java.util.*;
public class Solution {
ArrayList<Integer> list = new ArrayList<Integer>();
public ArrayList<Integer> preorderTraversal(TreeNode root) {
if(root == null)
return list;
Stack<TreeNode> stack = new Stack<TreeNode>();
stack.push(root);
while(!stack.isEmpty())
{
TreeNode te = stack.pop();
list.add(te.val);
if(te.right != null)
{
stack.push(te.right);
}
if(te.left != null)
{
stack.push(te.left);
}
}
return list;
}
}/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
//递归,先序遍历:根 左 右
import java.util.*;
public class Solution {
public ArrayList<Integer> preorderTraversal(TreeNode root) {
ArrayList<Integer> list = new ArrayList<Integer>();
if(root == null){ //判断root==null
return list; //{}返回为[]
}
list.add(root.val);
list.addAll(preorderTraversal(root.left)); //addAll:添加集合
list.addAll(preorderTraversal(root.right));
return list;
}
}
本文介绍了一种实现二叉树前序遍历的方法,包括迭代和递归两种方式。迭代方法使用栈来辅助完成节点值的收集,而递归方法则直接利用递归特性进行遍历。
161

被折叠的 条评论
为什么被折叠?



