题解 | #二叉树的中序遍历#

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[] inorderTraversal (TreeNode root) {
        // write code here
        ArrayList<Integer> list =new ArrayList<>();
        // inorder(root,list);
        // int[] res = new int[list.size()];
        // for(int i=0;i<res.length;i++){
        //     res[i] = list.get(i);
        // }
        // return res;
        
        return inorder2(root,list);
    }
    //递归写法
    public void inorder(TreeNode root,ArrayList<Integer> list){
        if(root==null) return;
        inorder(root.left,list);
        list.add(root.val);
        inorder(root.right,list);
    }
    //非递归写法
    public int[] inorder2(TreeNode root,ArrayList<Integer> list){
        if(root==null) return new int[]{};
        Stack<TreeNode> stack = new Stack<>();
        TreeNode node = root;
        while(node!=null||!stack.isEmpty()){
            //将左子树上的所有节点都压入栈中
            while(node!=null){
                stack.push(node);
                node =node.left;
            }
            node =stack.pop();
            list.add(node.val);
            //去右子树上看看有没有节点
            node = node.right;
        }
        return list.stream().mapToInt(Integer::intValue).toArray();
     }
}