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

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

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
     */


    public int[][] threeOrders (TreeNode root) {
        // write code here
        List<List<Integer>> lists = new ArrayList<>();
        for (int i = 0; i < 3; ++i)
            lists.add(new ArrayList<Integer>());
        getPreOrder(root, lists.get(0));
        getMidOrder(root, lists.get(1));
        getBackOrder(root, lists.get(2));
        int n = lists.get(0).size();        
        int [][] ans = new int[3][n];
        for (int i = 0; i < 3; ++i) {
            for (int j = 0; j < n; ++j) {
                ans[i][j] = lists.get(i).get(j);
            }
        }
        return ans;
    }

    // 前序
    List getPreOrder(TreeNode root, List list) {
        if (root == null) return null;
        list.add(root.val);//实际上不加泛型检查也可以
        getPreOrder(root.left, list);
        getPreOrder(root.right, list);
        return list;

    }

    //  中序
    List<Integer> getMidOrder(TreeNode root, List<Integer> list) {
        if (root == null) return null;
        getMidOrder(root.left, list);
        list.add(root.val);
        getMidOrder(root.right, list);
        return list;

    }

    // 后序
    List<Integer> getBackOrder(TreeNode root, List<Integer> list) {
        if (root == null) return null;
        getBackOrder(root.left, list);
        getBackOrder(root.right, list);
        list.add(root.val);
        return list;
    }
    
}

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03-10 20:35
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武汉大学 C++
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