第66题 Binary Tree Preorder Traversal

最新推荐文章于 2025-12-13 19:26:52 发布

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#stack #算法 #二叉树 #leetcode

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本文介绍了一种实现二叉树前序遍历的方法，包括递归和非递归两种方式。递归方法通过调用自身来遍历左右子树，而非递归方法则利用栈来跟踪节点。

Binary Tree Preorder Traversal

Description
Given a binary tree, return the preorder traversal of its nodes’ values.

法1：递归：

/**
 * 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: A Tree
     * @return: Preorder in ArrayList which contains node values.
     */
    public List<Integer> preorderTraversal(TreeNode root) {
        // write your code here
        ArrayList<Integer> results = new ArrayList<>() ;
        if (root == null){
           return results ;
        }
        List<Integer> left = preorderTraversal(root.left);
        List<Integer> right = preorderTraversal(root.right);
        results.add(root.val) ;
        results.addAll(left) ;
        results.addAll(right);
        return results ;
    }
}

法2：非递归：

/**
 * 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: A Tree
     * @return: Preorder in ArrayList which contains node values.
     */
    public List<Integer> preorderTraversal(TreeNode root) {
        // write your code here
      
        Stack<TreeNode> stack = new Stack<TreeNode>();
        ArrayList<Integer> results = new ArrayList<Integer>();
        if(root == null){
           return results ;
        }
        stack.push(root);
        while(!stack.isEmpty()){
           TreeNode node = stack.pop();
           results.add(node.val);
           if(node.right != null){
              stack.push(node.right) ;
           }
           if(node.left != null){
              stack.push(node.left);
           }
        }
        return results ;
    }
}