[数据结构与算法]144. 二叉树的前序遍历

144. 二叉树的前序遍历

难度简单610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]

递归

/**
 * 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) {
          List<Integer> res = new ArrayList<Integer>();
        return forward(res,root);
    }
    public static List<Integer> forward( List<Integer> res,TreeNode root){
        if(root==null) return res;
        res.add(root.val);
        forward(res,root.left);
        forward(res,root.right);
        return res;
    }
}

迭代

/**
 * 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) {
        List<Integer> res = new ArrayList<Integer>();
        Stack<TreeNode> stk = new Stack<TreeNode>();
        TreeNode p = root;
        while(p!=null||!stk.isEmpty()){
            while(p!=null){
                res.add(p.val);
                stk.add(p);
                p = p.left;
            }
            p = stk.pop();
            p = p.right;
        }
        return res;
    }
 
}