二叉树的递归遍历
递归的三要素
1.递归函数的参数和返回值
2.递归出口
3.单层递归的逻辑
144. 二叉树的前序遍历
/**
* 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 List<Integer> preorderTraversal(TreeNode root) {
List<Integer> result = new ArrayList<>();
preoder(root,result);
return result;
}
public void preoder(TreeNode node,List<Integer> result){
if (node==null){
return;
}
result.add(node.val);//前序遍历：中、左、右
preoder(node.left,result);
preoder(node.right,result);
}
}
94. 二叉树的中序遍历
/**
* 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 List<Integer> inorderTraversal(TreeNode root) {
List<Integer> result = new ArrayList();
inorder(root,result);
return result;
}
public void inorder(TreeNode node, List<Integer> result){
if(node==null){
return;
}
inorder(node.left,result);//中序遍历：左、中、右
result.add(node.val);
inorder(node.right,result);
}
}
145. 二叉树的后序遍历
/**
* 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 List<Integer> postorderTraversal(TreeNode root) {
List<Integer> result = new ArrayList();
postorder(root,result);
return result;
}
public void postorder(TreeNode node,List<Integer> result){
if(node==null){
return;
}
postorder(node.left,result);//后序遍历：左、右、中
postorder(node.right,result);
result.add(node.val);
}
}
二叉树的迭代遍历
用栈操作，递归也是用栈实现的嘛🙂
144. 二叉树的前序遍历
/**
* 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 List<Integer> preorderTraversal(TreeNode root) {
List<Integer> result = new ArrayList<>();//结果列表
Stack<TreeNode> stack = new Stack<>();
if(root==null){
return result;
}
stack.push(root);//先把根节点加到栈中去
while (!stack.empty()){
TreeNode node = stack.pop();//从栈中弹出一个结点来进行操作
result.add(node.val);//弹出的元素加到结果列表中
if(node.right!=null){
stack.push(node.right);//右孩子不空就进栈
}
if(node.left!=null){
stack.push(node.left);//左孩子不空就进栈
}
}
return result;
}
}
- 妙蛙种子吃了妙脆角，妙到家啦
145. 二叉树的后序遍历
/**
* 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 List<Integer> postorderTraversal(TreeNode root) {
List<Integer> result = new ArrayList<>();//结果列表
Stack<TreeNode> stack = new Stack<>();
if(root==null){
return result;
}
stack.push(root);//先把根节点加到栈中去
while (!stack.empty()){
TreeNode node = stack.pop();//从栈中弹出一个结点来进行操作
result.add(node.val);//弹出的元素加到结果列表中
if(node.left!=null){
stack.push(node.left);//左孩子不空就进栈
}
if(node.right!=null){
stack.push(node.right);//右孩子不空就进栈
}
}
Collections.reverse(result);
return result;
}
}
Collections.reverse(result) 链表反转
- 这题和前序遍历十分相似，就是入栈顺序不一样，画图找一下顺序，改前序遍历的代码就可啦
94. 二叉树的中序遍历
- 中序遍历和前序遍历、后续遍历不一样的地方是，前序遍历（中左右）、后序遍历（左右中），中结点在两端，处理结点就是当前遍历的结点（从根节点开始遍历，从根节点开始处理）。而中序遍历的遍历从根节点开始，要处理的结点却是从最左侧的结点开始。
/**
* 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 List<Integer> inorderTraversal(TreeNode root) {
List<Integer> result = new ArrayList<>();//结果列表
Stack<TreeNode> stack = new Stack<>();
if(root==null){
return result;
}
TreeNode cur = root;//取到根结点
while (cur != null || !stack.isEmpty()){
if (cur != null){
stack.push(cur);//放入栈中
cur = cur.left;//把当前结点的左孩子赋给当前结点
}else{
cur = stack.pop();//弹出栈中的结点
result.add(cur.val);//放入结果集中
cur = cur.right;//把当前结点的右孩子赋给当前结点（左边已经遍历完了，上一步也把中间放入结果集中，该右边了）
}
}
return result;
}
}
二叉树的层序遍历
也就是广度优先遍历啦
/**
* 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 List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> res = new ArrayList<>();//外层链表
Queue<TreeNode> que = new LinkedList<>();//新建一个队列
if (root == null) {
return res;
}
que.add(root);//把根节点放入队列
while (!que.isEmpty()){
//当队列不为空时
ArrayList<Integer> item = new ArrayList<>();//内层链表
int size = que.size();//队列的大小
while (size>0){
TreeNode node = que.poll();//弹出当前结点
if(node.left!=null){que.add(node.left);}//把当前结点的左孩子加进去（如果有的话）
if(node.right!=null){que.add(node.right);}//把当前结点的右孩子加进去（如果有的话）
item.add(node.val);//当前结点加到链表
size--;
}
res.add(item);//内层链表加入到外层链表中
}
return res;
}
}
下面是一堆层序遍历的题
107. 二叉树的层序遍历 II
/**
* 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 List<List<Integer>> levelOrderBottom(TreeNode root) {
List<List<Integer>> res = new ArrayList<>();//外层链表
Queue<TreeNode> que = new LinkedList<TreeNode>();//队列
if(root==null)return res;
que.add(root);//把根结点放入队列
while (!que.isEmpty()){
List<Integer> item = new ArrayList<>();
int size = que.size();
while (size > 0) {
TreeNode node = que.poll();//队列中弹出一个结点
item.add(node.val);
if(node.left!=null){que.add(node.left);}
if(node.right!=null){que.add(node.right);}
size--;
}
res.add(item);
}
Collections.reverse(res);
return res;
}
}
199. 二叉树的右视图
/**
* 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 List<Integer> rightSideView(TreeNode root) {
List<Integer> res = new ArrayList<>();
Queue<TreeNode> que = new LinkedList<>();
if(root==null)return res;
que.add(root);//根结点不为空，放入队列
while (!que.isEmpty()){
List<Integer> item = new ArrayList<>();
int size = que.size();
while (size>0){
TreeNode node = que.poll();
item.add(node.val);
if(node.left!=null){que.add(node.left);}
if(node.right!=null){que.add(node.right);}
size--;
}
Integer i = item.get(item.size() - 1);
res.add(i);
}
return res;
}
}
637. 二叉树的层平均值
/**
* 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 List<Double> averageOfLevels(TreeNode root) {
List<Double> res = new ArrayList<>();
Queue<TreeNode> que = new LinkedList<>();
if(root==null)return res;
que.add(root);
while (!que.isEmpty()){
int size = que.size();
double x = 0;
double sum = 0;
int count = size;
while (size>0){
TreeNode node = que.poll();
sum += node.val;
if(node.left!=null){que.add(node.left);}
if(node.right!=null){que.add(node.right);}
size--;
}
x = sum/count;
res.add(x);
}
return res;
}
}
429. N 叉树的层序遍历
/*
// 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 List<List<Integer>> levelOrder(Node root) {
List<List<Integer>> res = new ArrayList<>();//外层链表
Queue<Node> que = new LinkedList<>();//新建一个队列
if (root == null) {
return res;
}
que.add(root);//把根节点放入队列
while (!que.isEmpty()){
//当队列不为空时
ArrayList<Integer> item = new ArrayList<>();//内层链表
int size = que.size();//队列的大小
while (size>0){
Node node = que.poll();//弹出当前结点
//当前结点加到链表
if(node.children!=null){
for (Node child : node.children) {
que.add(child);
}
}
item.add(node.val);
size--;
}
res.add(item);//内层链表加入到外层链表中
}
return res;
}
}
- 添加子结点到队列的操作有点不一样
515. 在每个树行中找最大值
/**
* 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 List<Integer> largestValues(TreeNode root) {
List<Integer> res = new ArrayList<>();
Queue<TreeNode> que = new LinkedList<>();
if(root==null){
return res;
}
que.add(root);
while(!que.isEmpty()){
int size = que.size();
int x = Integer.MIN_VALUE;
while(size>0){
TreeNode node = que.poll();
x = node.val>x ? node.val : x;
if(node.left!=null){que.add(node.left);}
if(node.right!=null){que.add(node.right);}
size--;
}
res.add(x);
}
return res;
}
}
116. 填充每个节点的下一个右侧节点指针
/*
// Definition for a Node.
class Node {
public int val;
public Node left;
public Node right;
public Node next;
public Node() {}
public Node(int _val) {
val = _val;
}
public Node(int _val, Node _left, Node _right, Node _next) {
val = _val;
left = _left;
right = _right;
next = _next;
}
};
*/
class Solution {
public Node connect(Node root) {
Queue<Node> que = new LinkedList<>();
if (root == null) {
return root;
}
que.add(root);
while (que.size() > 0) {
int size = que.size();
Node node = que.poll();
if (node.left != null) {que.add(node.left);}
if (node.right != null) {que.add(node.right);}
for (int i = 1; i < size; i++) {
Node next = que.poll();//弹出该层剩余元素
if (next.left != null) que.add(next.left);
if (next.right != null) que.add(next.right);
node.next = next;
node = next;
}
}
return root;
}
}
117. 填充每个节点的下一个右侧节点指针 II
/*
// Definition for a Node.
class Node {
public int val;
public Node left;
public Node right;
public Node next;
public Node() {}
public Node(int _val) {
val = _val;
}
public Node(int _val, Node _left, Node _right, Node _next) {
val = _val;
left = _left;
right = _right;
next = _next;
}
};
*/
class Solution {
public Node connect(Node root) {
Queue<Node> que = new LinkedList<>();
if (root == null) {
return root;
}
que.add(root);
while (que.size() > 0) {
int size = que.size();
Node node = que.poll();
if (node.left != null) {que.add(node.left);}
if (node.right != null) {que.add(node.right);}
for (int i = 1; i < size; i++) {
Node next = que.poll();//弹出该层剩余元素
if (next.left != null) que.add(next.left);
if (next.right != null) que.add(next.right);
node.next = next;
node = next;
}
}
return root;
}
}
- 离大谱，这题代码跟上题一样，一模一样
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) {
Queue<TreeNode> que = new LinkedList<>();//新建一个队列
if (root == null) {
return 0;
}
que.add(root);//把根节点放入队列
int count = 0;
while (!que.isEmpty()) {
//当队列不为空时
count++;
ArrayList<Integer> item = new ArrayList<>();//内层链表
int size = que.size();//队列的大小
while (size > 0) {
TreeNode node = que.poll();//弹出当前结点
if (node.left != null) {
que.add(node.left);
}//把当前结点的左孩子加进去（如果有的话）
if (node.right != null) {
que.add(node.right);
}//把当前结点的右孩子加进去（如果有的话）
size--;
}
}
return count;
}
}
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) {
Queue<TreeNode> que = new LinkedList<>();//新建一个队列
if (root == null) {
return 0;
}
que.add(root);//把根节点放入队列
int count = 0;
while (!que.isEmpty()) {
//当队列不为空时
count++;
ArrayList<Integer> item = new ArrayList<>();//内层链表
int size = que.size();//队列的大小
while (size > 0) {
TreeNode node = que.poll();//弹出当前结点
if (node.left != null) {
que.add(node.left);
}//把当前结点的左孩子加进去（如果有的话）
if (node.right != null) {
que.add(node.right);
}//把当前结点的右孩子加进去（如果有的话）
if(node.left==null&&node.right==null){
return count;
}
size--;
}
}
return count;
}
}
总结
通过今天的题目大致把二叉树的结构掌握。深度优先遍历方面掌握前序、中序、后续的递归实现和迭代实现。掌握广度优先遍历的模板（写了十道层序遍历的题目，就算是小猪也会了😐
今天的题目自己写出来的不多，除了最后几道改模板的题，不知道是因为天太冷还是头上戴的蝴蝶结封印了我的智慧的🙃