代码随想录算法训练营20期|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
104.二叉树的最大深度
递归法：
/**
* 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 left = maxDepth(root.left);
int right = maxDepth(root.right);
return Math.max(left, right) + 1;
}
}迭代法
/**
* 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 level = 0;
Queue queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
int size = queue.size();
level++;
for (int i = 0; i < size; i++) {
TreeNode cur = queue.poll();
if (cur.left != null) queue.add(cur.left);
if (cur.right != null) queue.add(cur.right);
}
}
return level;
}
} 559.n叉树的最大深度
递归：
/*
// Definition for a Node.
class Node {
public int val;
public List children;
public Node() {}
public Node(int _val) {
val = _val;
}
public Node(int _val, List _children) {
val = _val;
children = _children;
}
};
*/
class Solution {
public int maxDepth(Node root) {
if (root == null) return 0;
int depth = 0;
for (Node child : root.children) {
if (child != null) {
depth = Math.max(depth, maxDepth(child));
}
}
return depth + 1;
}
} 迭代法
/*
// Definition for a Node.
class Node {
public int val;
public List children;
public Node() {}
public Node(int _val) {
val = _val;
}
public Node(int _val, List _children) {
val = _val;
children = _children;
}
};
*/
class Solution {
public int maxDepth(Node root) {
if (root == null) return 0;
int depth = 0;
Queue queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
int size = queue.size();
depth++;
for (int i = 0; i < size; i++) {
Node cur = queue.poll();
if (cur.children != null) {
for (Node child : cur.children) {
queue.add(child);
}
}
}
}
return depth;
}
} 111.二叉树的最小深度
递归法
/**
* 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;
int left = minDepth(root.left);
int right = minDepth(root.right);
if (root.left == null && root.right != null) {
return right + 1;
}
if (root.left != null && root.right == null) {
return left + 1;
}
int res = Math.min(left, right) + 1;
return res;
}
}迭代法
/**
* 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 queue = new LinkedList<>();
queue.add(root);
int depth = 0;
while (!queue.isEmpty()) {
int size = queue.size();
depth++;
for (int i = 0; i < size; i++) {
TreeNode cur = queue.poll();
if (cur.left == null && cur.right == null) {
return depth;
}
if (cur.left != null) queue.add(cur.left);
if (cur.right != null) queue.add(cur.right);
}
}
return depth;
}
} 222.完全二叉树的节点个数
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 left = countNodes(root.left);
int right = countNodes(root.right);
int res = left + right + 1;
return res;
}
}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) {
if (root == null) return 0;
int count = 0;
Queue queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; i++) {
TreeNode cur = queue.poll();
count++;
if (cur.left != null) queue.add(cur.left);
if (cur.right != null) queue.add(cur.right);
}
}
return count;
}
}