算法练习第十六天| 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

优先掌握递归的方式

104.二叉树的最大深度

/*** 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 Solution {// public int maxDepth(TreeNode root) {//     if(root == null ) return 0;//     int depth = 0;//     Queue<TreeNode> queue = new LinkedList<>();//     queue.offer(root);//     while(!queue.isEmpty()){//         int len = queue.size();//         depth++;//         while(len > 0){//             TreeNode tmpNode = queue.poll();//             if(tmpNode.left != null) queue.offer(tmpNode.left);//             if(tmpNode.right != null) queue.offer(tmpNode.right);//             len--;//         }//     }//     return depth;// }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;}
}

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 Solution {// public int maxDepth(Node root) {//     if(root==null) return 0;//     Queue<Node> queue = new LinkedList<>();//     queue.offer(root);//     int depth = 0;//     while(!queue.isEmpty()){//         depth++;//         int len = queue.size();//         while(len>0){//             Node node = queue.poll();//             for(int i = 0;i<node.children.size(); i++){//                 if(node.children.get(i)!= null){//                     queue.offer(node.children.get(i));//                 }//             }//             len--;//         }//     }//     return depth;// }public int maxDepth(Node root) {if(root==null) return 0;int depth = 0;if(root.children != null){for(Node child : root.children){depth = Math.max(depth,maxDepth(child));}}return depth+1;}
}

111.二叉树的最小深度

/*** 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 Solution {// public int minDepth(TreeNode root) {//     if(root== null) return 0;//     Queue<TreeNode> queue = new LinkedList<>();//     queue.offer(root);//     int depth = 0;//     while(!queue.isEmpty()){//         depth ++;//         int size = queue.size();//         for(int i = 0; i<size;i++){//             TreeNode tmpNode = queue.poll();//             if(tmpNode.left == null && tmpNode.right==null)//             return depth;//             if(tmpNode.left != null) queue.offer(tmpNode.left);//             if(tmpNode.right != null) queue.offer(tmpNode.right);//         }//     }//     return depth;// }public int minDepth(TreeNode root) {if(root== null) return 0;int leftDepth = minDepth(root.left);int rightDepth = minDepth(root.right);if(root.left != null && root.right == null){return leftDepth + 1;}if(root.right != null && root.left == null){return rightDepth + 1;}return Math.min(leftDepth,rightDepth) +  1;}
}

222.完全二叉树的节点个数

/*** 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 leftCount = countNodes(root.left);//     int rightCount = countNodes(root.right);//     return leftCount + rightCount +1;// }public int countNodes(TreeNode root) {if(root==null) return 0;Queue<TreeNode> queue = new LinkedList<>();int res = 0;queue.offer(root);while(!queue.isEmpty()){int size = queue.size();while(size-- >0){TreeNode node = queue.poll();res++;if(node.left != null) queue.offer(node.left);if(node.right != null) queue.offer(node.right);}}return res;}
}