界面设计是做什么的,windows优化大师的特点,做网站游燕,网站设计建设那家好提示#xff1a;文章写完后#xff0c;目录可以自动生成#xff0c;如何生成可参考右边的帮助文档 文章目录 前言一、力扣104. 二叉树的最大深度二、力扣559. N 叉树的最大深度三、力扣111. 二叉树的最小深度三、力扣力扣222. 完全二叉树的节点个数 前言 一、力扣104. 二叉树… 提示文章写完后目录可以自动生成如何生成可参考右边的帮助文档 文章目录 前言一、力扣104. 二叉树的最大深度二、力扣559. N 叉树的最大深度三、力扣111. 二叉树的最小深度三、力扣力扣222. 完全二叉树的节点个数 前言 一、力扣104. 二叉树的最大深度
递归
/*** 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 Solution {public int maxDepth(TreeNode root) {if(root null){return 0;}int l maxDepth(root.left);int r maxDepth(root.right);return l r ? l 1 : r 1;}
}迭代
/*** 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 Solution {public int maxDepth(TreeNode root) {DequeTreeNode deq new LinkedList();if(root null)return 0;deq.offerLast(root);int high 0;while(!deq.isEmpty()){int len deq.size();for(int i 0; i len; i ){TreeNode p deq.pollFirst();if(p.left! null)deq.offerLast(p.left);if(p.right ! null)deq.offerLast(p.right);}high ;}return high;}
}二、力扣559. N 叉树的最大深度
迭代
/*
// Definition for a Node.
class Node {public int val;public ListNode children;public Node() {}public Node(int _val) {val _val;}public Node(int _val, ListNode _children) {val _val;children _children;}
};
*/class Solution {public int maxDepth(Node root) {DequeNode deq new LinkedList();if(root null)return 0;deq.offerLast(root);int high 0;while(!deq.isEmpty()){int len deq.size();for(int i 0; i len; i ){Node p deq.pollFirst();ListNode li p.children;for(Node n : li){if(n ! null){deq.offerLast(n);}}}high ;}return high;}
}递归
/*
// Definition for a Node.
class Node {public int val;public ListNode children;public Node() {}public Node(int _val) {val _val;}public Node(int _val, ListNode _children) {val _val;children _children;}
};
*/class Solution {public int maxDepth(Node root) {if(root null)return 0;int[] arr new int[root.children.size()];int max 0;for(int i 0; i arr.length; i ){arr[i] maxDepth(root.children.get(i));max max arr[i] ? max : arr[i];}return max 1;}
}三、力扣111. 二叉树的最小深度
迭代
/*** 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 Solution {public int minDepth(TreeNode root) {DequeTreeNode deq new LinkedList();if(root null)return 0;deq.offerLast(root);int depth 0;while(!deq.isEmpty()){int len deq.size();for(int i 0; i len ; i ){TreeNode p deq.pollFirst();if(p.left null p.right null){return depth 1;}if(p.left ! null)deq.offerLast(p.left);if(p.right ! null)deq.offerLast(p.right);}depth ;}return depth;}
}递归
class Solution {/*** 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;}
}三、力扣力扣222. 完全二叉树的节点个数
迭代
/*** 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 Solution {public int countNodes(TreeNode root) {DequeTreeNode deq new LinkedList();if(root null)return 0;deq.offerLast(root);int count 0;while(!deq.isEmpty()){int len deq.size();for(int i 0; i len ; i ){TreeNode p deq.pollFirst();count ;if(p.left ! null)deq.offerLast(p.left);if(p.right ! null)deq.offerLast(p.right);}}return count;}
}递归
/*** 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 Solution {public int countNodes(TreeNode root) {if(root null)return 0;int l countNodes(root.left);int r countNodes(root.right);return l r 1;}
}
