Leedcode算法专题训练（树）
2024-09-07 04:37:16
递归
一棵树要么是空树，要么有两个指针，每个指针指向一棵树。树是一种递归结构，很多树的问题可以使用递归来处理。
1. 树的高度
104. Maximum Depth of Binary Tree (Easy)
class Solution {
public int maxDepth(TreeNode root) {
if(root==null)return 0;
return Math.max(maxDepth(root.left),maxDepth(root.right))+1;
}
}2. 平衡树
110. Balanced Binary Tree (Easy)
3
/ \
9 20
/ \
15 7平衡树左右子树高度差都小于等于 1
class Solution {
public boolean result=true;
public boolean isBalanced(TreeNode root) {
if(root==null)return true;
if(Math.abs(depth(root.left)-depth(root.right))>1)return false;
return isBalanced(root.left)&&isBalanced(root.right);
}
public int depth(TreeNode root){
if(root==null)return 0;
return Math.max(depth(root.left),depth(root.right))+1;
}
}3. 两节点的最长路径
543. Diameter of Binary Tree (Easy)
Input:
1
/ \
2 3
/ \
4 5
Return 3, which is the length of the path [4,2,1,3] or [5,2,1,3].class Solution {
private int max = 0;
public int diameterOfBinaryTree(TreeNode root) {
depth(root);
return max;
}
private int depth(TreeNode root) {
if (root == null) return 0;
int leftDepth = depth(root.left);
int rightDepth = depth(root.right);
max = Math.max(max, leftDepth + rightDepth);
return Math.max(leftDepth, rightDepth) + 1;
}
}4. 翻转树
226. Invert Binary Tree (Easy)
class Solution {
public TreeNode invertTree(TreeNode root) {
if(root==null)return null;
TreeNode tmp=root.left;
root.left=root.right;
root.right=tmp;
invertTree(root.left);
invertTree(root.right);
return root;
}
}5. 归并两棵树
617. Merge Two Binary Trees (Easy)
Input:
Tree 1 Tree 2
1 2
/ \ / \
3 2 1 3
/ \ \
5 4 7
Output:
3
/ \
4 5
/ \ \
5 4 7class Solution {
public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
if(t1==null&&t2==null)return null;
if(t1==null)return t2;
if(t2==null)return t1;
TreeNode root =new TreeNode(t1.val+t2.val);
root.left=mergeTrees(t1.left,t2.left);
root.right=mergeTrees(t1.right,t2.right);
return root;
}
}6. 判断路径和是否等于一个数
Leetcdoe : 112. Path Sum (Easy)
Given the below binary tree and sum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ \
7 2 1
return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.class Solution {
public boolean hasPathSum(TreeNode root, int sum) {
if(root==null)return false;
if(root.left == null && root.right == null){
return root.val == sum;
}
return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
}
}class Solution {
public List<List<Integer>> pathSum(TreeNode root, int sum) {
List<List<Integer>> result=new ArrayList<>();
dfs(root,sum,new ArrayList<>(),result);
return result;
}
private void dfs(TreeNode root,int sum,List<Integer> list,List<List<Integer>>result){
if(root==null){
return;
}
list.add(root.val);
if(root.left==null&&root.right==null&&sum==root.val){
result.add(new ArrayList(list));
}
dfs(root.left,sum-root.val,list,result);
dfs(root.right,sum-root.val,list,result);
list.remove(list.size() - 1);
}
}
}7. 统计路径和等于一个数的路径数量
437. Path Sum III (Easy)
java 双重递归 思路：首先先序递归遍历每个节点，再以每个节点作为起始点递归寻找满足条件的路径。
class Solution {
public int pathSum(TreeNode root, int sum) {
if(root==null)return 0;
int result=countPath(root,sum);
int a=pathSum(root.left,sum);
int b=pathSum(root.right,sum);
return result+a+b;
}
public int countPath(TreeNode root,int sum){
if(root==null){
return 0;
}
sum=sum-root.val;
int result=sum==0?1:0;
return result+countPath(root.left,sum)+countPath(root.right,sum);
}
}8. 子树
572. Subtree of Another Tree (Easy)
class Solution {
public boolean isSubtree(TreeNode s, TreeNode t) {
if (s == null) return false;
return isSubtreeWithRoot(s, t) || isSubtree(s.left, t) || isSubtree(s.right, t);
}
private boolean isSubtreeWithRoot(TreeNode s, TreeNode t) {
if(t==null&&s==null)return true;
if(t==null||s==null)return false;
if(t.val != s.val)return false;
return isSubtreeWithRoot(s.left, t.left) && isSubtreeWithRoot(s.right, t.right);
}
}9. 树的对称
101. Symmetric Tree (Easy)
1
/ \
2 2
/ \ / \
3 4 4 3class Solution {
public boolean isSymmetric(TreeNode root) {
if(root==null)return true;
return isSymmetrictree(root.left,root.right);
}
private boolean isSymmetrictree(TreeNode left,TreeNode right){
if(left==null&&right==null)return true;
if(left==null||right==null)return false;
if(left.val!=right.val)return false;
return isSymmetrictree(left.left,right.right)&&isSymmetrictree(left.right,right.left);
}
}10. 最小路径
111. Minimum Depth of Binary Tree (Easy)
树的根节点到叶子节点的最小路径长度
public int minDepth(TreeNode root) {
if (root == null) return 0;
int left = minDepth(root.left);
int right = minDepth(root.right);
if (left == 0 || right == 0) return left + right + 1;
return Math.min(left, right) + 1;
}11. 统计左叶子节点的和
404. Sum of Left Leaves (Easy)
解题思路
1、终止条件,root == null时，返回0;
2、如果当前节点是左节点，并且无子节点，累加对应节点的val值;
3、继续搜索左右子树，直到root == null，即最后一个节点。
class Solution {
private int res=0;
public int sumOfLeftLeaves(TreeNode root) {
if(root==null){
return 0;
}
if(root.left!=null&&root.left.left==null&&root.left.right==null){
res+=root.left.val;
}
sumOfLeftLeaves(root.left);
sumOfLeftLeaves(root.right);
return res;
}
}12. 相同节点值的最大路径长度
687. Longest Univalue Path (Easy)
1
/ \
4 5
/ \ \
4 4 5
Output : 2class Solution {
int ans;
public int longestUnivaluePath(TreeNode root) {
ans = 0;
longestPath(root);
return ans;
}
//递归函数功能：搜寻以node为起点的最长同值路径:要么是以node为起点的左子树，要么是以node为起点的右子树
public int longestPath(TreeNode node) {
if (node == null) return 0;
int maxLorRres=0;
int left = longestPath(node.left); //node左子树的最长同值路径
int right = longestPath(node.right);//node右子树的最长同值路径
//这种情况对于寻找最长同值路径长有帮助，对于搜索以root为路径起始点的最长路径没有帮助
if (node.left != null && node.left.val == node.val&&node.right != null && node.right.val == node.val) {
ans=Math.max(ans, left + right+2);
}
//从左右子树中选择最长的同值路径
if(node.left!=null&&node.left.val == node.val){
maxLorRres=left+1;
}
if(node.right!=null&&node.right.val==node.val){
maxLorRres=Math.max(maxLorRres,right+1);
}
//从ans与maxLorRres中更新最大值
ans=Math.max(ans,maxLorRres);
return maxLorRres; //所以你能知道为什么返回这个值了吗？
}
}private int path = 0;
public int longestUnivaluePath(TreeNode root) {
dfs(root);
return path;
}
private int dfs(TreeNode root){
if (root == null) return 0;
int left = dfs(root.left);
int right = dfs(root.right);
int leftPath = root.left != null && root.left.val == root.val ? left + 1 : 0;
int rightPath = root.right != null && root.right.val == root.val ? right + 1 : 0;
path = Math.max(path, leftPath + rightPath);
return Math.max(leftPath, rightPath);
}13. 间隔遍历
337. House Robber III (Medium)
3
/ \
2 3
\ \
3 1
Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.class Solution {
public int rob(TreeNode root) {
if(root==null)return 0;
int val1=root.val;
if(root.left!=null)val1+=rob(root.left.left)+rob(root.left.right);
if(root.right!=null)val1+=rob(root.right.left) + rob(root.right.right);
int val2 = rob(root.left) + rob(root.right);
return Math.max(val1, val2);
}
}14. 找出二叉树中第二小的节点
671. Second Minimum Node In a Binary Tree (Easy)
Input:
2
/ \
2 5
/ \
5 7
Output: 5class Solution {
public int findSecondMinimumValue(TreeNode root) {
return traversal(root,root.val);
}
private int traversal(TreeNode root,int value){
if(root == null){
return -1;
}
if(root.val > value){
return root.val;
}
// 寻找左右子节点中，第一个大于自己的节点
int l = traversal(root.left,value);
int r = traversal(root.right,value);
// 存在两个子节点
if(l>=0 && r>=0){
return Math.min(l,r);
}
// 存在0个子节点
return Math.max(l,r);
}
}层次遍历
使用 BFS 进行层次遍历。不需要使用两个队列来分别存储当前层的节点和下一层的节点，因为在开始遍历一层的节点时，当前队列中的节点数就是当前层的节点数，只要控制遍历这么多节点数，就能保证这次遍历的都是当前层的节点。
1. 一棵树每层节点的平均数
637. Average of Levels in Binary Tree (Easy)
class Solution {
public List<Double> averageOfLevels(TreeNode root) {
List<Double> result=new ArrayList<>();
if(root==null)return result;
Queue<TreeNode>queue=new LinkedList<>();
queue.add(root);
while(!queue.isEmpty()){
double sum=0;
int length=queue.size();
for(int i=queue.size();i>0;i--){
TreeNode temp=queue.poll();
sum+=temp.val;
if(temp.left!=null)queue.add(temp.left);
if(temp.right!=null)queue.add(temp.right);
}
result.add(sum/length);
}
return result;
}
}2. 得到左下角的节点
513. Find Bottom Left Tree Value (Easy)
Input:
1
/ \
2 3
/ / \
4 5 6
/
7
Output:
7class Solution {
public int findBottomLeftValue(TreeNode root) {
if(root==null)return 0;
Queue<TreeNode> queue=new LinkedList<>();
queue.add(root);
int result=0;
while(!queue.isEmpty()){
TreeNode temp=queue.poll();
result=temp.val;
if (temp.right != null) queue.add(temp.right);
if (temp.left != null) queue.add(temp.left);
}
return result;
}
}前中后序遍历
1
/ \
2 3
/ \ \
4 5 6- 层次遍历顺序：[1 2 3 4 5 6]
- 前序遍历顺序：[1 2 4 5 3 6]
- 中序遍历顺序：[4 2 5 1 3 6]
- 后序遍历顺序：[4 5 2 6 3 1]
层次遍历使用 BFS 实现，利用的就是 BFS 一层一层遍历的特性；而前序、中序、后序遍历利用了 DFS 实现。
前序、中序、后序遍只是在对节点访问的顺序有一点不同，其它都相同。
① 前序
void dfs(TreeNode root) {
visit(root);
dfs(root.left);
dfs(root.right);
}② 中序
void dfs(TreeNode root) {
dfs(root.left);
visit(root);
dfs(root.right);
}③ 后序
void dfs(TreeNode root) {
dfs(root.left);
dfs(root.right);
visit(root);
}1. 非递归实现二叉树的前序遍历
144. Binary Tree Preorder Traversal (Medium)
class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> list=new ArrayList<>();
Stack<TreeNode> stack=new Stack<>();
while(root!=null||(!stack.empty())){
if(root!=null){
list.add(root.val);//步骤一，取根节点的值
stack.push(root);//把根节点放入栈中
root=root.left;//步骤二，遍历左子树
}
else{
TreeNode tem=stack.pop();
root=tem.right;//步骤三，遍历右子树
}
}
return list;
}
}2. 非递归实现二叉树的中序遍历
94. Binary Tree Inorder Traversal (Medium)
class Solution {
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> list=new ArrayList<>();
Stack<TreeNode> stack=new Stack<>();
while(root!=null||(!stack.empty())){
if(root!=null){
stack.push(root);//把根节点放入栈中
root=root.left;//步骤一，遍历左子树
}
else{
TreeNode tem=stack.pop();
list.add(tem.val);//步骤二，取根结点的值
root=tem.right;//步骤三，遍历右子树
}
}
return list;
}
}3. 非递归实现二叉树的后序遍历
145. Binary Tree Postorder Traversal (Medium)
class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> list=new ArrayList<>();
Stack<TreeNode> stack=new Stack<>();
while(root!=null||(!stack.empty())){
if(root!=null){
stack.push(root);//把根节点放入栈中
list.add(0,root.val);//步骤一，在index=0处插入根结点的值
root=root.right;//步骤二，遍历右子树
}
else{
TreeNode tem=stack.pop();
root=tem.left;//步骤三，遍历左子树
}
}
return list;
}
}BST
二叉查找树（BST）：根节点大于等于左子树所有节点，小于等于右子树所有节点。
二叉查找树中序遍历有序。
1. 修剪二叉查找树
669. Trim a Binary Search Tree (Easy)
class Solution {
public TreeNode trimBST(TreeNode root, int low, int high) {
if(root==null)return root;
if(root.val>high)return trimBST(root.left,low,high);
if(root.val<low)return trimBST(root.right,low,high);
root.left=trimBST(root.left,low,high);
root.right=trimBST(root.right,low,high);
return root;
}
}2. 寻找二叉查找树的第 k 个元素
230. Kth Smallest Element in a BST (Medium)
中序遍历解法：
class Solution {
private int val;
private int cnt=0;
public int kthSmallest(TreeNode root, int k) {
inOrder(root,k);
return val;
}
private void inOrder(TreeNode node,int k){
if(node==null)return;
inOrder(node.left,k);
cnt++;
if(cnt==k){
val=node.val;
return;
}
inOrder(node.right,k);
}
}class Solution {
public int kthSmallest(TreeNode root, int k) {
Stack<TreeNode>stack=new Stack<>();
while(true){
while(root!=null){
stack.add(root);
root=root.left;
}
root=stack.pop();
if(--k==0)return root.val;
root=root.right;
}
}
}3. 把二叉查找树每个节点的值都加上比它大的节点的值
Convert BST to Greater Tree (Easy)
Input: The root of a Binary Search Tree like this:
5
/ \
2 13
Output: The root of a Greater Tree like this:
18
/ \
20 13class Solution {
int sum=0;
public TreeNode convertBST(TreeNode root) {
if(root==null)return null;
convertBST(root.right);
sum+=root.val;
root.val=sum;
convertBST(root.left);
return root;
}
}4. 二叉查找树的最近公共祖先
235. Lowest Common Ancestor of a Binary Search Tree (Easy)
class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if(root==null)return null;
if(root.val>p.val&&root.val>q.val)return lowestCommonAncestor(root.left,p,q);
else if(root.val<p.val&&root.val<q.val)return lowestCommonAncestor(root.right,p,q);
return root;
}
}5. 二叉树的最近公共祖先
236. Lowest Common Ancestor of a Binary Tree (Medium)
class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if(root==null||root==p||root==q)return root;
TreeNode left=lowestCommonAncestor(root.left,p,q);
TreeNode right=lowestCommonAncestor(root.right,p,q);
if(left!=null&&right!=null)return root;
else if(left!=null&&right==null)return left;
else return right;
}
}6. 从有序数组中构造二叉查找树
108. Convert Sorted Array to Binary Search Tree (Easy)
class Solution {
public TreeNode sortedArrayToBST(int[] nums) {
return helper(nums,0,nums.length-1);
}
public TreeNode helper(int[] nums,int lo,int hi){
if(lo>hi){
return null;
}
int mid=lo+(hi-lo)/2;
TreeNode root=new TreeNode(nums[mid]);
root.left=helper(nums,lo,mid-1);
root.right=helper(nums,mid+1,hi);
return root;
}
}7. 根据有序链表构造平衡的二叉查找树
109. Convert Sorted List to Binary Search Tree (Medium)
class Solution {
public TreeNode sortedListToBST(ListNode head) {
if(head==null)return null;
if(head.next==null)return new TreeNode(head.val);
ListNode pre=null,fast=head,slow=head;
while(fast!=null&&fast.next!=null){
pre=slow;
slow=slow.next;
fast=fast.next.next;
}
pre.next=null;
TreeNode root=new TreeNode(slow.val);
root.left=sortedListToBST(head);
root.right=sortedListToBST(slow.next);
return root;
}
}8. 在二叉查找树中寻找两个节点，使它们的和为一个给定值
653. Two Sum IV - Input is a BST (Easy)
public class Solution {
public boolean findTarget(TreeNode root, int k) {
Set < Integer > set = new HashSet();
return find(root, k, set);
}
public boolean find(TreeNode root, int k, Set < Integer > set) {
if (root == null)
return false;
if (set.contains(k - root.val))
return true;
set.add(root.val);
return find(root.left, k, set) || find(root.right, k, set);
}
}
9. 在二叉查找树中查找两个节点之差的最小绝对值
这题的难点在于思考前向节点如何表示，还有BST的特性不难
530. Minimum Absolute Difference in BST (Easy)
class Solution {
int abs_min=Integer.MAX_VALUE;
TreeNode pre;
public int getMinimumDifference(TreeNode root) {
if(root==null)return 0;
inOrder(root);
return abs_min;
}
private void inOrder(TreeNode node){
if(node==null)return;
inOrder(node.left);
if(pre!=null) abs_min=Math.min(abs_min,node.val-pre.val);
pre=node;
inOrder(node.right);
}
}10. 寻找二叉查找树中出现次数最多的值
501. Find Mode in Binary Search Tree (Easy)
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
int preVal = 0, curTimes = 0, maxTimes = 0;
ArrayList<Integer> list = new ArrayList<Integer>();
public int[] findMode(TreeNode root) {
traversal(root);
//list转换为int[]
int size = list.size();
int[] ans = new int[size];
for(int i = 0; i < size; i++){
ans[i] = list.get(i);
}
return ans;
}
//二叉搜索树中序遍历是递增顺序
public void traversal(TreeNode root){
if(root != null){
traversal(root.left);
//判断当前值与上一个值的关系, 更新 curTimes 和 preVal
if(preVal == root.val){
curTimes++;
}else{
preVal = root.val;
curTimes = 1;
}
//判断当前数量与最大数量的关系, 更新 list 和 maxTimes
if(curTimes == maxTimes){
list.add(root.val);
}else if(curTimes > maxTimes){
list.clear();
list.add(root.val);
maxTimes = curTimes;
}
traversal(root.right);
}
}
}Trie
Trie，又称前缀树或字典树，用于判断字符串是否存在或者是否具有某种字符串前缀。
1. 实现一个 Trie
208. Implement Trie (Prefix Tree) (Medium)
这个题目的关键是数组的用法，就是说建立一个对象，对象中的数组进行实例化，不断进入对象的孩子数组中再进行实例化
class TrieNode{
TrieNode[] child;
boolean isEnd;
public TrieNode(){
child=new TrieNode[26];
isEnd=false;
}
}
class Trie {
TrieNode root;
/** Initialize your data structure here. */
public Trie() {
root=new TrieNode();
}
/** Inserts a word into the trie. */
public void insert(String word) {
TrieNode cur=root;
for(int i=0;i<word.length();i++){
char c=word.charAt(i);
if(cur.child[c-'a']==null){
cur.child[c-'a']=new TrieNode();
}
cur=cur.child[c-'a'];
}
cur.isEnd=true;
}
/** Returns if the word is in the trie. */
public boolean search(String word) {
TrieNode cur=root;
for(int i=0;i<word.length();i++){
char c=word.charAt(i);
if(cur.child[c-'a']==null){
return false;
}
cur=cur.child[c-'a'];
}
return cur.isEnd;
}
/** Returns if there is any word in the trie that starts with the given prefix. */
public boolean startsWith(String prefix) {
TrieNode cur=root;
for(int i=0;i<prefix.length();i++){
char c=prefix.charAt(i);
if(cur.child[c-'a']==null){
return false;
}
cur=cur.child[c-'a'];
}
return true;
}
}hashmap的是实现
class TrieNode{
HashMap<Character,TrieNode> next=new HashMap<>();
boolean isEnd;
public TrieNode(){
isEnd=false;
}
}
class Trie {
TrieNode root;
/** Initialize your data structure here. */
public Trie() {
root=new TrieNode();
}
/** Inserts a word into the trie. */
public void insert(String word) {
TrieNode cur=root;
for(int i=0;i<word.length();i++){
char c=word.charAt(i);
if(!cur.next.containsKey(c)){
cur.next.put(c,new TrieNode());
}
cur=cur.next.get(c);
}
cur.isEnd=true;
}
/** Returns if the word is in the trie. */
public boolean search(String word) {
TrieNode cur=root;
for(int i=0;i<word.length();i++){
char c=word.charAt(i);
if(!cur.next.containsKey(c)){
return false;
}
cur=cur.next.get(c);
}
return cur.isEnd;
}
/** Returns if there is any word in the trie that starts with the given prefix. */
public boolean startsWith(String prefix) {
TrieNode cur=root;
for(int i=0;i<prefix.length();i++){
char c=prefix.charAt(i);
if(!cur.next.containsKey(c)){
return false;
}
cur=cur.next.get(c);
}
return true;
}
}2. 实现一个 Trie，用来求前缀和
677. Map Sum Pairs (Medium)
class MapSum {
class TrieNode{
TrieNode[] children;
int value;
public TrieNode(){
children=new TrieNode[26];
}
}
private TrieNode root;
/** Initialize your data structure here. */
public MapSum() {
root=new TrieNode();
}
public void insert(String key, int val) {
TrieNode node=root;
char[] charArray=key.toCharArray();
for(int i=0;i<charArray.length;i++){
int idx=charArray[i]-'a';
if(node.children[idx]==null){
node.children[idx]=new TrieNode();
}
node=node.children[idx];
}
node.value=val;
}
public int sum(String prefix) {
if(prefix==null)return 0;
TrieNode node=root;
char[] charArray=prefix.toCharArray();
for(int i=0;i<charArray.length;i++){
int idx=charArray[i]-'a';
if(node.children[idx]==null)return 0;
node=node.children[idx];
}
return dfs(node);
}
private int dfs(TrieNode node){
if(node==null)return 0;
int res=node.value;
for(TrieNode child:node.children){
res+=dfs(child);
}
return res ;
}
}
/**
* Your MapSum object will be instantiated and called as such:
* MapSum obj = new MapSum();
* obj.insert(key,val);
* int param_2 = obj.sum(prefix);
*/最新文章
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