题解 | 实现二叉树先序，中序和后序遍历

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 the root of binary tree
     * @return int整型二维数组
     */
    List<Integer> list = new ArrayList<>();
    public int[][] threeOrders (TreeNode root) {
        // write code here
        preorder(root, list);
        int[][] res = new int[3][list.size()];
        for (int i=0; i<list.size(); i++) {
            res[0][i] = list.get(i);
        }
        list.clear();
        inorder(root, list);
        for (int i=0; i<list.size();i++) {
            res[1][i] = list.get(i);
        }
        list.clear();
        postorder(root, list);
        for (int i=0; i<list.size();i++) {
            res[2][i] = list.get(i);
        }

        return res;
    }


    private void preorder(TreeNode root, List<Integer> list) {
        if (root == null) return;
        Stack<TreeNode> sk = new Stack<>();
        sk.push(root);
        while (!sk.isEmpty()) {
            TreeNode node = sk.pop();
            if (node.right != null) {
                sk.push(node.right);
            }
            if (node.left != null) {
                sk.push(node.left);
            }
            list.add(node.val);
        }
        //System.out.println(list);   
    }

    private void inorder(TreeNode root, List<Integer> list) {
        Stack<TreeNode> sk = new Stack<>();
        while (!sk.isEmpty() || root != null) {
            while (root != null) {
                sk.push(root);
                root = root.left;
            }
            TreeNode cur = sk.pop();
            list.add(cur.val);
            if (cur.right != null) {
                root = cur.right;
            }
        }
    }

    private void postorder(TreeNode root, List<Integer> list) {
        Stack<TreeNode> sk = new Stack<>();
        TreeNode pre = null;
        TreeNode cur = root;
        while (!sk.isEmpty() || cur != null) {
            while (cur != null) {
                sk.push(cur);
                cur = cur.left;
            }
            cur = sk.peek();

            if(cur.right == null || cur.right == pre) { // 如果右子树为空 或者当前节点右节点 之前访问过
                sk.pop();
                list.add(cur.val);
                pre = cur;
                cur = null;
            } else {
                cur = cur.right;
            }  

        }
    }
}