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;
 *   }
 * }
 */

/**
 * NC45 实现二叉树先序，中序和后序遍历
 * @author d3y1
 */
public class Solution {
    private ArrayList<Integer> list = new ArrayList<>();

    /**
     * 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
     */
    public int[][] threeOrders (TreeNode root) {
        return solution1(root);
        // return solution2(root);
    }

    /**
     * 递归法
     * @param root
     * @return
     */
    private int[][] solution1(TreeNode root){
        run(root);

        int n = list.size()/3;
        int[][] results = new int[3][n];
        for(int i=0; i<3; i++){
            for(int j=0; j<n; j++){
                results[i][j] = list.get(i*n+j);
            }
        }

        return results;
    }

    /**
     * 按序运行: 前序 -> 中序 -> 后序
     * @param root
     */
    private void run(TreeNode root){
        preOrder(root);
        inOrder(root);
        postOrder(root);
    }

    /**
     * 前序遍历
     * @param root
     */
    private void preOrder(TreeNode root){
        if(root == null){
            return;
        }

        list.add(root.val);
        preOrder(root.left);
        preOrder(root.right);
    }

    /**
     * 中序遍历
     * @param root
     */
    private void inOrder(TreeNode root){
        if(root == null){
            return;
        }

        inOrder(root.left);
        list.add(root.val);
        inOrder(root.right);
    }

    /**
     * 后序遍历
     * @param root
     */
    private void postOrder(TreeNode root){
        if(root == null){
            return;
        }

        postOrder(root.left);
        postOrder(root.right);
        list.add(root.val);
    }

    //////////////////////////////////////////////////////////////////////////////////////

    /**
     * 迭代法
     * @param root
     * @return
     */
    private int[][] solution2(TreeNode root){
        operate(root);

        int n = list.size()/3;
        int[][] results = new int[3][n];
        for(int i=0; i<3; i++){
            for(int j=0; j<n; j++){
                results[i][j] = list.get(i*n+j);
            }
        }

        return results;
    }

    /**
     * 按序运行: 前序 -> 中序 -> 后序
     * @param root
     */
    private void operate(TreeNode root){
        preorder(root);
        inorder(root);
        postorder(root);
    }

    /**
     * 前序遍历: 栈
     * @param root
     * @return
     */
    private void preorder(TreeNode root){
        Stack<TreeNode> stack = new Stack<>();

        if(root == null){
            return;
        }

        stack.push(root);

        TreeNode node;
        while(!stack.isEmpty()){
            node = stack.pop();
            list.add(node.val);
            if(node.right != null){
                stack.push(node.right);
            }
            if(node.left != null){
                stack.push(node.left);
            }
        }
    }

    /**
     * 中序遍历: 栈
     * @param root
     * @return
     */
    private void inorder(TreeNode root){
        Stack<TreeNode> stack = new Stack<>();

        TreeNode curr;
        while(root!=null || !stack.isEmpty()){
            // 最左边
            while(root != null){
                stack.push(root);
                root = root.left;
            }

            curr = stack.pop();
            list.add(curr.val);

            root = curr.right;
        }
    }

    /**
     * 后序遍历: 栈
     * @param root
     * @return
     */
    private void postorder(TreeNode root){
        Stack<TreeNode> stack = new Stack<>();

        TreeNode curr;
        // 上次访问
        TreeNode last = null;
        while(root!=null || !stack.isEmpty()){
            // 最左边
            while(root != null){
                stack.push(root);
                root = root.left;
            }

            curr = stack.pop();
            // 右节点为空或已访问
            if(curr.right==null || curr.right==last){
                // 访问当前节点
                list.add(curr.val);
                last = curr;
            }
            // 右节点非空且未访问
            else{
                // 当前节点再次入栈
                stack.push(curr);
                root = curr.right;
            }
        }
    }
}