题解 | #二叉树的前序遍历#TOP23

思路:
1.简单的前序遍历 ,根 左右
2.利用栈实现BFS，或者递归

import java.util.*;

/*
 * public class TreeNode {
 *   int val = 0;
 *   TreeNode left = null;
 *   TreeNode right = null;
 *   public TreeNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param root TreeNode类 
     * @return int整型一维数组
     */
    public int[] preorderTraversal (TreeNode root) {
        // write code here
        if(root == null ){
            return new int[]{};
        }
        //根左右
        ArrayList<Integer> result = new ArrayList<>();
        preOrderBFS(root, result);
        return result.stream().mapToInt(Integer::intValue).toArray();
        
    }
    private void preOrder(TreeNode node ,ArrayList<Integer> result){
        if(node == null){
            return;
        }
        result.add(node.val);
        preOrder(node.left, result);
        preOrder(node.right, result);
    }
    private void preOrderBFS(TreeNode node ,ArrayList<Integer> result){
        if(node == null){
            return;
        }
        Stack<TreeNode> stack = new Stack<>();
        stack.push(node);
        
        while(!stack.empty()){
            TreeNode temp = stack.pop();
            result.add(temp.val);
            if(temp.right != null){
                stack.push(temp.right);
            }
            if(temp.left != null){
                stack.push(temp.left);
            }
        }
        
    }
}