题解 | #实现二叉树先序,中序和后序遍历#

实现二叉树先序,中序和后序遍历

http://www.nowcoder.com/practice/a9fec6c46a684ad5a3abd4e365a9d362

import java.util.*;
import java.util.*;
/*
 * public class TreeNode {
 *   int val = 0;
 *   TreeNode left = null;
 *   TreeNode right = null;
 * }
 */

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) {
        // write code here
        if(root == null){
            return new int[3][0];
        }
        ArrayList<Integer> list1 = preOrder(root);
        ArrayList<Integer> list2 = inOrder(root);
        ArrayList<Integer> list3 = postOrder(root);
        int[][] res = new int[3][list1.size()];
        for(int i = 0; i < list1.size(); ++i){
            res[0][i] = list1.get(i);
        }

        for(int i = 0; i < list2.size(); ++i){
            res[1][i] = list2.get(i);
        }

        for(int i = 0; i < list3.size(); ++i){
            res[2][i] = list3.get(i);
        }

        return res;
    }

    public ArrayList<Integer> preOrder(TreeNode root){
        ArrayList<Integer> res = new ArrayList<>();

        res.add(root.val);
        if(root.left != null){
            res.addAll(preOrder(root.left));
        }
        if(root.right != null){
            res.addAll(preOrder(root.right));
        }
        return res;
    }

    public ArrayList<Integer> inOrder(TreeNode root){
        ArrayList<Integer> res = new ArrayList<>();

        if(root.left != null){
            res.addAll(inOrder(root.left));
        }

        res.add(root.val);

        if(root.right != null){
            res.addAll(inOrder(root.right));
        }
        return res;
    }

    public ArrayList<Integer> postOrder(TreeNode root){
        ArrayList<Integer> res = new ArrayList<>();

        if(root.left != null){
            res.addAll(postOrder(root.left));
        }

        if(root.right != null){
            res.addAll(postOrder(root.right));
        }

        res.add(root.val);

        return res;
    }
}
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