关于算法:算法tree二叉树的最大深度

题目地址

我的解法

    /**
     * 2ms 21.07
     * 41M 33.17
     * 其实是广度优先搜寻
     * 切入点是一层一层看，依据以后层构建下一层。* */
    public static int maxDepth(TreeNode root) {if (null == root) {return 0;}
        int deep = 1;
        List<TreeNode> curLevel = new ArrayList<>();
        curLevel.add(root);
        List<TreeNode> nextLevel = null;
        while (hasNextLevel(curLevel)) {nextLevel = new ArrayList<>();
            for (TreeNode cur : curLevel) {if (null != cur) {nextLevel.add(cur.left);
                    nextLevel.add(cur.right);
                }
            }
            curLevel = nextLevel;
            deep ++;
        }
        return deep;
    }

官网 - 广度优先搜寻

    /**
     * 官网的广度优先搜寻
     * 使用了 Queue @see MyQueue
     * */
    public static int maxDepth_l1(TreeNode root) {if (null == root) {return 0;}

        Queue<TreeNode> curLevel = new LinkedList<>();
        curLevel.offer(root);

        int deep = 0;
        while (!curLevel.isEmpty()) {int size = curLevel.size();//TODO 固定 poll 次数
            while (size > 0) {TreeNode curNode = curLevel.poll();
                if (curNode == null) {continue;}
                if (curNode.left != null) {curLevel.offer(curNode.left);
                }
                if (curNode.right != null) {curLevel.offer(curNode.right);
                }
                size --;
            }

//            Queue<TreeNode> curLevelTemp  = curLevel;
//            for (TreeNode cur : curLevelTemp) {java.util.ConcurrentModificationException
//                if (cur == null) {
//                    continue;
//                }
//                if (cur.left != null) {//                    curLevel.offer(cur.left);
//                }
//                if (cur.right != null) {//                    curLevel.offer(cur.right);
//                }
//                curLevel.poll();
//            }
            deep ++;
        }
        return deep;
    }

官网 - 深度优先搜寻

    /**
     * 官网深度优先搜寻
     * 使用递归
     * 如果咱们晓得了左子树和右子树的最大深度 l 和 r，那么该二叉树的最大深度即为
     *
     * max(l,r) + 1
     *
     * 而左子树和右子树的最大深度又能够以同样的形式进行计算。* */
    public static int maxDepth_l2(TreeNode root) {
        //    3
        //   / \
        //  9  20
        //    /  \
        //   15   7
        if (root == null) {return 0;} else {int maxLeft = maxDepth_l2(root.left);//1
            int maxRight = maxDepth_l2(root.right);
            return Math.max(maxLeft, maxRight) + 1;
        }
    }

拓展 -DFS&BFS

深度优先搜索算法 DFS

1、算法步骤：
——拜访顶点 v；
——顺次从 v 的未被拜访的邻接点登程，对图进行深度优先遍历；直至图中和 v 有门路相通的顶点都被拜访；
——若此时图中尚有顶点未被拜访，则从一个未被拜访的顶点登程，从新进行深度优先遍历，直到图中所有顶点均被拜访过为止
2、DFS 属于自觉搜寻。深度优先搜寻是图论中的经典算法，利用深度优先搜索算法能够产生指标图的相应拓扑排序表，利用拓扑排序表能够不便的解决很多相干的图论问题，如最大门路问题等等。个别用堆数据结构来辅助实现 DFS 算法。

广度优先搜索算法 BFS

1、算法步骤：
——首先将根节点放入队列中；
——从队列中取出第一个节点，并测验它是否为指标。如果找到指标，则完结搜查并回传后果。否则将它所有尚未测验过的间接子节点退出队列中；
——若队列为空，示意整张图都查看过了——亦即图中没有欲搜查的指标。完结搜查并回传“找不到指标”；
——反复步骤 2
2、BFS 同样属于自觉搜寻。简略的说，BFS 是从根节点开始，沿着树 (图) 的宽度遍历树 (图) 的节点。如果所有节点均被拜访，则算法停止。个别用队列数据结构来辅助实现 BFS 算法。

拓展 API-Queue

个性

1、线性表 / 线性队列
2、FIFO
3、表头出，表尾加

implements

        Queue<String> queue_ArrayBlockingQueue = new ArrayBlockingQueue<>(10);
        Queue<String> queue_ArrayDeque = new ArrayDeque<>();
        Queue<String> queue_ConcurrentLinkedDeque = new ConcurrentLinkedDeque<>();
        Queue<String> queue_LinkedBlockingDeque = new LinkedBlockingDeque<>();
        Queue<String> queue_LinkedList = new LinkedList<>();

methods

code

    public static void main(String[] args) {Queue<String> queue_ArrayBlockingQueue = new ArrayBlockingQueue<>(10);
        Queue<String> queue_ArrayDeque = new ArrayDeque<>();
        Queue<String> queue_ConcurrentLinkedDeque = new ConcurrentLinkedDeque<>();
        Queue<String> queue_LinkedBlockingDeque = new LinkedBlockingDeque<>();
        Queue<String> queue_LinkedList = new LinkedList<>();
        System.out.println("++++++++++++++++++++++++++");

        //add 表尾插 ——>[1,2,3]
        // 队列满时，抛出 IllegalStateException
        Queue<Integer> queue_add = new ArrayBlockingQueue<>(2);
        queue_add.add(1);
        queue_add.add(2);
        try {queue_add.add(3);
        } catch (Exception e) {System.out.println(e);
        }
        for (Integer q : queue_add) {System.out.println(q);
        }

        try {Queue<Integer> queue_LinkedList_add = new LinkedList<>();
            for (int i = 0; i < 1000000; i++) {queue_LinkedList_add.add(i);
            }
            System.out.println(queue_LinkedList_add.size());
        } catch (Exception e) {System.out.println(e);
        }

        System.out.println("++++++++++++++++++++++++++");

        //offer 表尾插 ——>[1,2,3,4,5,6]
        // 队列满时，返回 false
        Queue<Integer> queue_offer = new ArrayBlockingQueue<>(4);
        System.out.println(queue_offer.offer(1));
        System.out.println(queue_offer.offer(2));
        System.out.println(queue_offer.offer(3));
        System.out.println(queue_offer.offer(4));
        try {System.out.println(queue_offer.offer(5));
        } catch (Exception e) {System.out.println(e);
        }
        for (Integer q : queue_offer) {System.out.println(q);
        }

        try {Queue<Integer> queue_LinkedList_offer = new LinkedList<>();
            for (int i = 0; i < 1000000; i++) {if (!queue_LinkedList_offer.add(i)) {System.out.println("offer false");
                }
            }
            System.out.println(queue_LinkedList_offer.size());
        } catch (Exception e) {System.out.println(e);
        }
        System.out.println("++++++++++++++++++++++++++");

        //remove 移除并返回队列头部元素
        // 队列空时，抛出 NoSuchElementException
        Queue<Integer> queue_LinkedList_remove = new LinkedList<>();
        for (int i = 0; i < 6; i++) {queue_LinkedList_remove.offer(i);
        }
        for (int i = 0; i < 7; i++) {
            try {System.out.println(queue_LinkedList_remove.remove());
            } catch (Exception e) {System.out.println(i + "e =" + e);
            }
        }
        System.out.println("++++++++++++++++++++++++++");

        //poll 移除并返回队列头部元素
        // 队列空时，返回 null
        Queue<Integer> queue_poll = new LinkedList<>();
        for (int i = 0; i < 10; i++) {queue_poll.offer(i);
        }
        for (int i = 0; i < 11; i++) {
            try {System.out.println(queue_poll.poll());
            } catch (Exception e) {System.out.println(i + "e =" + e);
            }
        }
        System.out.println("++++++++++++++++++++++++++");


        //element 返回队列头部元素
        // 队列空时，抛出 NoSuchElementException
        Queue<Integer> queue_element = new LinkedList<>();
        queue_element.offer(1);
        queue_element.offer(2);
        System.out.println(queue_element.element());
        System.out.println(queue_element.poll());
        System.out.println(queue_element.element());
        System.out.println(queue_element.poll());
        try {System.out.println(queue_element.element());
        } catch (Exception e) {System.out.println(e);
        }
        System.out.println("++++++++++++++++++++++++++");


        //peek 返回队列头部元素
        // 队列空时，返回 null
        Queue<Integer> queue_peek = new LinkedList<>();
        for (int i = 0; i < 3; i++) {queue_peek.offer(i);
        }
        for (int i = 0; i < 4; i++) {
            try {System.out.println(queue_peek.peek());
                System.out.println(queue_peek.poll());
            } catch (Exception e) {System.out.println(i + "e =" + e);
            }
        }
        System.out.println("++++++++++++++++++++++++++");
    }