LeetCode-面试算法经典-Java实现107-Binary Tree Level Order Traversal II（二叉树层序遍历II）

Posted 2020-09-29 zhchoutai

【107-Binary Tree Level Order Traversal II（二叉树层序遍历II）】

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【LeetCode-面试算法经典-Java实现】【全部题目文件夹索引】

原题

　　Given a binary tree, return the bottom-up level order traversal of its nodes’ values. (ie, from left to right, level by level from leaf to root).
　　For example:
　　Given binary tree {3,9,20,#,#,15,7},

    3
   /   9  20
    /     15   7

　　return its bottom-up level order traversal as:

[
  [15,7],
  [9,20],
  [3]
]

题目大意

　　给定一棵二叉树自底向下进行层序遍历。

解题思路

　　对树进行层序遍历，每层的结果放在结果链表的头部。

代码实现

public class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode(int x) { val = x; }
}

算法实现类

import java.util.*;

public class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {

        List<List<Integer>> list = new LinkedList<>();
        if (root == null) {
            return list;
        }

        Deque<TreeNode> cur = new LinkedList<>();
        Deque<TreeNode> nxt = new LinkedList<>();
        Deque<TreeNode> exc = new LinkedList<>();

        TreeNode tmp;

        cur.add(root);
        while (!cur.isEmpty()) {
            List<Integer> layout = new ArrayList<>();

            while (!cur.isEmpty()) {
                tmp = cur.remove();

                if (tmp.left != null) {
                    nxt.add(tmp.left);
                }

                if (tmp.right != null) {
                    nxt.add(tmp.right);
                }

                layout.add(tmp.val);
            }

            exc = cur;
            cur = nxt;
            nxt = exc;

            list.add(0, layout);
        }

        return list;
    }
}

评測结果

　　点击图片，鼠标不释放，拖动一段位置。释放后在新的窗体中查看完整图片。

特别说明

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