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("++++++++++++++++++++++++++");
}
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