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
| method | 个性 |
|---|---|
| add | 队列尾插入，队列满时抛出IllegalStateException |
| offer | 队列尾插入，队列满时返回false |
| remove | 移除并返回头部元素，队列空时，抛出NoSuchElementException |
| pool | 移除并返回头部元素，队列空时，返回null |
| element | 查看队列头元素，队列空时，抛出NoSuchElementException |
| peek | 查看队列头元素，队列空时，返回null |
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("++++++++++++++++++++++++++"); }