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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) {
Deque<TreeNode> 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 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) {
Deque<Node> 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();
List<Node> 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 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;
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) {
Deque<TreeNode> 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null
return Math.min(leftDepth, rightDepth) + 1;
}
}
三、力扣力扣222. 完全二叉树的节点个数
迭代文章来源：https://www.toymoban.com/news/detail-697994.html
/**
* 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) {
Deque<TreeNode> 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;
}
}
递归文章来源地址https://www.toymoban.com/news/detail-697994.html
/**
* 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;
}
}
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