【每日一题】【list转int数组】【Lambda的简化-方法引用】2022年1月15日-NC45 实现二叉树先序，中序和后序遍历

描述

给定一棵二叉树，分别按照二叉树先序，中序和后序打印所有的节点。

 

数据范围：0 \le n \le 10000≤n≤1000，树上每个节点的val值满足 0 \le val \le 1000≤val≤100

要求：空间复杂度 O(n)O(n)，时间复杂度 O(n)O(n)

样例解释：

答案：Lambda的简化-方法引用，list转int数组

import java.util.*;
public class Solution {
    /**
     * 
     * @param root TreeNode类 the root of binary tree
     * @return int整型二维数组
     */
    
    public int[][] threeOrders (TreeNode root) {
        List<Integer> preList = new ArrayList<>();
        List<Integer> inList = new ArrayList<>();
        List<Integer> postList = new ArrayList<>();
        preOrder(root, preList);
        inOrder(root, inList);
        postOrder(root, postList);
        int[][] res = new int[3][preList.size()];
        //转为int数组，需要用mapToInt
        res[0] = preList.stream().mapToInt(Integer::valueOf).toArray();
        res[1] = inList.stream().mapToInt(Integer::valueOf).toArray();
        res[2] = postList.stream().mapToInt(Integer::valueOf).toArray();
        return res;
    }
    
    public void preOrder(TreeNode root, List<Integer> preList) {
        if(root == null) return;
        preList.add(root.val);
        preOrder(root.left, preList);
        preOrder(root.right, preList);
    }
    
    public void inOrder(TreeNode root, List<Integer> inList) {
        if(root == null) return;
        inOrder(root.left, inList);
        inList.add(root.val);
        inOrder(root.right, inList);
    }
    
    public void postOrder(TreeNode root, List<Integer> postList) {
        if(root == null) return;
        postOrder(root.left, postList);
        postOrder(root.right, postList);
        postList.add(root.val);
    }
}