代码随想录-算法训练营day16【二叉树03：二叉树的最大深度、二叉树的最小深度、完全二叉树的节点个数】

代码随想录-035期-算法训练营【博客笔记汇总表】-CSDN博客

第六章  二叉树part03今日内容： ● 104.二叉树的最大深度  559.n叉树的最大深度
● 111.二叉树的最小深度
● 222.完全二叉树的节点个数迭代法，大家可以直接过，二刷有精力的时候 再去掌握迭代法。详细布置 104.二叉树的最大深度 （优先掌握递归）什么是深度，什么是高度，如何求深度，如何求高度，这里有关系到二叉树的遍历方式。大家 要先看视频讲解，就知道以上我说的内容了，很多录友刷过这道题，但理解的还不够。题目链接/文章讲解/视频讲解： https://programmercarl.com/0104.%E4%BA%8C%E5%8F%89%E6%A0%91%E7%9A%84%E6%9C%80%E5%A4%A7%E6%B7%B1%E5%BA%A6.html  111.二叉树的最小深度 （优先掌握递归）先看视频讲解，和最大深度 看似差不多，其实 差距还挺大，有坑。题目链接/文章讲解/视频讲解：https://programmercarl.com/0111.%E4%BA%8C%E5%8F%89%E6%A0%91%E7%9A%84%E6%9C%80%E5%B0%8F%E6%B7%B1%E5%BA%A6.html 222.完全二叉树的节点个数（优先掌握递归）需要了解，普通二叉树 怎么求，完全二叉树又怎么求题目链接/文章讲解/视频讲解：https://programmercarl.com/0222.%E5%AE%8C%E5%85%A8%E4%BA%8C%E5%8F%89%E6%A0%91%E7%9A%84%E8%8A%82%E7%82%B9%E4%B8%AA%E6%95%B0.html  往日任务
● day 1 任务以及具体安排：https://docs.qq.com/doc/DUG9UR2ZUc3BjRUdY  
● day 2 任务以及具体安排：https://docs.qq.com/doc/DUGRwWXNOVEpyaVpG  
● day 3 任务以及具体安排：https://docs.qq.com/doc/DUGdqYWNYeGhlaVR6 
● day 4 任务以及具体安排：https://docs.qq.com/doc/DUFNjYUxYRHRVWklp 
● day 5 周日休息
● day 6 任务以及具体安排：https://docs.qq.com/doc/DUEtFSGdreWRuR2p4 
● day 7 任务以及具体安排：https://docs.qq.com/doc/DUElCb1NyTVpXa0Jj 
● day 8 任务以及具体安排：https://docs.qq.com/doc/DUGdsY2JFaFhDRVZH 
● day 9 任务以及具体安排：https://docs.qq.com/doc/DUHVXSnZNaXpVUHN4 
● day 10 任务以及具体安排：https://docs.qq.com/doc/DUElqeHh3cndDbW1Q 
●day 11 任务以及具体安排：https://docs.qq.com/doc/DUHh6UE5hUUZOZUd0 
●day 12 周日休息 
●day 13 任务以及具体安排：https://docs.qq.com/doc/DUHNpa3F4b2dMUWJ3 
●day 14 任务以及具体安排：https://docs.qq.com/doc/DUHRtdXZZSWFkeGdE 
●day 15 任务以及具体安排：https://docs.qq.com/doc/DUHN0ZVJuRmVYeWNv

0104_二叉树的最大深度

0559_n叉树的最大深度

0111_二叉树的最小深度

0222_完全二叉树的节点个数

0104_二叉树的最大深度

package com.question.solve.leetcode.programmerCarl2._07_binaryTrees;import java.util.Deque;
import java.util.LinkedList;public class _0104_二叉树的最大深度 {
}/*** Definition for a binary tree node.* public class TreeNode {* int val;* TreeNode left;* TreeNode right;* TreeNode() {}* TreeNode(int val) { this.val = val; }* TreeNode(int val, TreeNode left, TreeNode right) {* this.val = val;* this.left = left;* this.right = right;* }* }*/
class Solution0104 {public int maxDepth(TreeNode root) {int depth = 0;if (root == null) {return depth;}Deque<TreeNode> deque = new LinkedList<TreeNode>();deque.offer(root);while (!deque.isEmpty()) {int size = deque.size();for (int i = 0; i < size; i++) {TreeNode poll = deque.poll();if (poll.left != null) {deque.offer(poll.left);}if (poll.right != null) {deque.offer(poll.right);}}depth++;}return depth;}public int maxDepth2(TreeNode root) {
//        if (root == null) return 0;
//        return 1 + Math.max(maxDepth(root.left), maxDepth(root.right));int depth = 0;if (root == null) {return depth;}int left = maxDepth2(root.left);int right = maxDepth2(root.right);depth = 1 + Math.max(left, right);return depth;}
}class Solution0104_2 {/*** 递归法*/public int maxDepth(TreeNode root) {if (root == null) {return 0;}int leftDepth = maxDepth(root.left);int rightDepth = maxDepth(root.right);return Math.max(leftDepth, rightDepth) + 1;}/*** 迭代法，使用层序遍历*/public int maxDepth2(TreeNode root) {if (root == null) {return 0;}Deque<TreeNode> deque = new LinkedList<>();deque.offer(root);int depth = 0;while (!deque.isEmpty()) {int size = deque.size();depth++;for (int i = 0; i < size; i++) {TreeNode node = deque.poll();if (node.left != null) {deque.offer(node.left);}if (node.right != null) {deque.offer(node.right);}}}return depth;}
}class Solution0104_3 {/*** 递归法(求深度法)*/int maxnum = 0;//定义最大深度public int maxDepth(TreeNode root) {ans(root, 0);return maxnum;}//递归求解最大深度void ans(TreeNode tr, int tmp) {if (tr == null) return;tmp++;maxnum = maxnum < tmp ? tmp : maxnum;ans(tr.left, tmp);ans(tr.right, tmp);tmp--;}
}

0559_n叉树的最大深度

package com.question.solve.leetcode.programmerCarl2._07_binaryTrees;import java.util.Deque;
import java.util.LinkedList;
import java.util.Queue;public class _0559_N叉树的最大深度 {
}/*
// Definition for a Node.
class Node {public int val;public List<Node> children;public Node() {}public Node(int _val) {val = _val;}public Node(int _val, List<Node> _children) {val = _val;children = _children;}
};
*/class Solution0559 {public int maxDepth(Node2 root) {if (root == null) {return 0;}Deque<Node2> deque = new LinkedList<>();deque.offer(root);int depth = 0;while (!deque.isEmpty()) {int size = deque.size();for (int i = 0; i < size; i++) {Node2 poll = deque.poll();if (poll.children != null) {for (Node2 x : poll.children) {deque.offer(x);}}}depth++;}return depth;}
}class Solution0559_2 {public int maxDepth(Node2 root) {return getDepth(root);}public int getDepth(Node2 node) {int depth = 0;if (node == null) {return depth;}for (Node2 x : node.children) {int temp = getDepth(x);if (depth < temp) {depth = temp;}}return depth + 1;}
}class Solution0559_3 {/*递归法，后序遍历求root节点的高度*/public int maxDepth(Node2 root) {if (root == null) return 0;int depth = 0;if (root.children != null) {for (Node2 child : root.children) {depth = Math.max(depth, maxDepth(child));}}return depth + 1; //中节点}/*** 迭代法，使用层序遍历*/public int maxDepth2(Node2 root) {if (root == null) return 0;int depth = 0;Queue<Node2> que = new LinkedList<>();que.offer(root);while (!que.isEmpty()) {depth++;int len = que.size();while (len > 0) {Node2 node = que.poll();for (int i = 0; i < node.children.size(); i++)if (node.children.get(i) != null)que.offer(node.children.get(i));len--;}}return depth;}
}

0111_二叉树的最小深度

package com.question.solve.leetcode.programmerCarl2._07_binaryTrees;import java.util.Deque;
import java.util.LinkedList;public class _0111_二叉树的最小深度 {
}/*** Definition for a binary tree node.* public class TreeNode {* int val;* TreeNode left;* TreeNode right;* TreeNode() {}* TreeNode(int val) { this.val = val; }* TreeNode(int val, TreeNode left, TreeNode right) {* this.val = val;* this.left = left;* this.right = right;* }* }*/
class Solution0111 {/*** 递归法，相比求MaxDepth要复杂点* 因为最小深度是从根节点到最近**叶子节点**的最短路径上的节点数量*/public int minDepth(TreeNode root) {if (root == null) {return 0;}int leftDepth = minDepth(root.left);int rightDepth = minDepth(root.right);if (root.left == null) {return rightDepth + 1;}if (root.right == null) {return leftDepth + 1;}// 左右结点都不为nullreturn Math.min(leftDepth, rightDepth) + 1;}
}class Solution0111_2 {/*** 递归法（思路来自二叉树最大深度的递归法）* 该题求最小深度，最小深度为根节点到叶子节点的深度，所以在迭代到每个叶子节点时更新最小值。*/int depth = 0;int minDepth = Integer.MAX_VALUE;//定义最小深度，初始化最大值public int minDepth(TreeNode root) {dep(root);return minDepth == Integer.MAX_VALUE ? 0 : minDepth;}void dep(TreeNode root) {if (root == null) return;//递归开始，深度增加depth++;dep(root.left);dep(root.right);//该位置表示递归到叶子节点了，需要更新最小深度minDepthif (root.left == null && root.right == null)minDepth = Math.min(minDepth, depth);//递归结束，深度减小depth--;}
}class Solution0111_3 {/*** 迭代法，层序遍历*/public int minDepth(TreeNode root) {if (root == null) {return 0;}Deque<TreeNode> deque = new LinkedList<>();deque.offer(root);int depth = 0;while (!deque.isEmpty()) {int size = deque.size();depth++;for (int i = 0; i < size; i++) {TreeNode poll = deque.poll();if (poll.left == null && poll.right == null) {//是叶子结点，直接返回depth，因为从上往下遍历，所以该值就是最小值return depth;}if (poll.left != null) {deque.offer(poll.left);}if (poll.right != null) {deque.offer(poll.right);}}}return depth;}
}

0222_完全二叉树的节点个数

package com.question.solve.leetcode.programmerCarl2._07_binaryTrees;import java.util.Deque;
import java.util.LinkedList;
import java.util.Queue;public class _0222_完全二叉树的节点个数 {
}class Solution0222 {public int countNodes(TreeNode root) {//通用递归解法if (root == null) {return 0;}return countNodes(root.left) + countNodes(root.right) + 1;}public int countNodes2(TreeNode root) {//迭代法if (root == null) return 0;Queue<TreeNode> queue = new LinkedList<>();queue.offer(root);int result = 0;while (!queue.isEmpty()) {int size = queue.size();while (size-- > 0) {TreeNode cur = queue.poll();result++;if (cur.left != null) queue.offer(cur.left);if (cur.right != null) queue.offer(cur.right);}}return result;}public int countNodes3(TreeNode root) {//二叉树层序遍历模板if (root == null) {return 0;}Deque<TreeNode> deque = new LinkedList<>();deque.offer(root);int num = 1;while (!deque.isEmpty()) {int size = deque.size();for (int i = 0; i < size; i++) {TreeNode poll = deque.poll();if (poll.left != null) {deque.offer(poll.left);num++;}if (poll.right != null) {deque.offer(poll.right);num++;}}}return num;}/*** 针对完全二叉树的解法* 满二叉树的结点数为：2^depth - 1*/public int countNodes4(TreeNode root) {if (root == null) return 0;TreeNode left = root.left;TreeNode right = root.right;int leftDepth = 0, rightDepth = 0; //这里初始为0是有目的的，为了下面求指数方便while (left != null) {  //求左子树深度left = left.left;leftDepth++;}while (right != null) { //求右子树深度right = right.right;rightDepth++;}if (leftDepth == rightDepth) {return (2 << leftDepth) - 1; //注意(2<<1) 相当于2^2，所以leftDepth初始为0}return countNodes(root.left) + countNodes(root.right) + 1;}
}