力扣https://leetcode-cn.com/problems/binary-tree-preorder-traversal/
难度简单725收藏分享切换为英文接收动态反馈
给你二叉树的根节点 root
,返回它节点值的 前序 遍历。
示例 1:
输入:root = [1,null,2,3] 输出:[1,2,3]
示例 2:
输入:root = [] 输出:[]
示例 3:
输入:root = [1] 输出:[1]
示例 4:
输入:root = [1,2] 输出:[1,2]
示例 5:
输入:root = [1,null,2] 输出:[1,2]
提示:
- 树中节点数目在范围
[0, 100]
内 -100 <= Node.val <= 100
递归法:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
List<Integer> ans = new ArrayList<Integer>();
public List<Integer> preorderTraversal(TreeNode root) {
pre(root);
return ans;
}
public void pre(TreeNode node)
{
if(node==null) return;
ans.add(node.val);
pre(node.left);
pre(node.right);
}
}
迭代法:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
// 非递归方法:迭代
List<Integer> ans = new ArrayList<Integer>();
// 创建栈
Stack<TreeNode> stack = new Stack<>();
if(root!=null) stack.push(root);
while(!stack.isEmpty())
{
TreeNode node = stack.pop();
ans.add(node.val);
if(node.right!=null) stack.push(node.right);
if(node.left!=null) stack.push(node.left);
}
return ans;
}
}