leecode更新

Leecode/src/main/java/com/markilue/leecode/tree/DeleteNode.java

@@ -4,7 +4,7 @@ package com.markilue.leecode.tree;
 
 import java.util.List;
 
-import static com.markilue.leecode.tree.InorderTraversal.inorderTraversal1;
+import static com.markilue.leecode.tree.T03_InorderTraversal.inorderTraversal1;
 
 /**
  * 删除二叉排序树的某个节点

Leecode/src/main/java/com/markilue/leecode/tree/DeleteNodeII.java

@@ -4,7 +4,7 @@ import org.junit.Test;
 
 import java.util.List;
 
-import static com.markilue.leecode.tree.InorderTraversal.inorderTraversal1;
+import static com.markilue.leecode.tree.T03_InorderTraversal.inorderTraversal1;
 
 /**
  * @BelongsProject: Leecode

Leecode/src/main/java/com/markilue/leecode/tree/T02_PostOrderTraversal.java

@@ -13,7 +13,7 @@ import java.util.*;
  * 给你二叉树的根节点 root ，返回它节点值的 前序 遍历。
  * @Version: 1.0
  */
-public class PostOrderTraversal {
+public class T02_PostOrderTraversal {
 
 
     @Test

Leecode/src/main/java/com/markilue/leecode/tree/T03_InorderTraversal.java

@@ -8,7 +8,7 @@ import java.util.Stack;
 /**
  * 中序遍历二叉排序树
  */
-public class InorderTraversal {
+public class T03_InorderTraversal {
 
     public static void main(String[] args) {
 

Leecode/src/main/java/com/markilue/leecode/tree/T04_LevelOrder.java

@@ -14,7 +14,7 @@ import java.util.*;
  * 给你二叉树的根节点 root ，返回其节点值的 层序遍历 。 （即逐层地，从左到右访问所有节点）。
  * @Version: 1.0
  */
-public class LevelOrder {
+public class T04_LevelOrder {
 
     @Test
     public void test() {

Leecode/src/main/java/com/markilue/leecode/tree/TrimBST.java

@@ -2,11 +2,9 @@ package com.markilue.leecode.tree;
 
 import org.junit.Test;
 
-import java.util.HashMap;
 import java.util.List;
-import java.util.Stack;
 
-import static com.markilue.leecode.tree.InorderTraversal.inorderTraversal1;
+import static com.markilue.leecode.tree.T03_InorderTraversal.inorderTraversal1;
 
 /**
  * @BelongsProject: Leecode

Leecode/src/main/java/com/markilue/leecode/tree/second/T02_PostOrderTraversal.java

@@ -0,0 +1,220 @@
+package com.markilue.leecode.tree.second;
+
+import com.markilue.leecode.tree.TreeNode;
+import com.markilue.leecode.tree.TreeUtils;
+import org.junit.Test;
+
+import java.util.*;
+
+/**
+ *@BelongsProject: Leecode
+ *@BelongsPackage: com.markilue.leecode.tree.second
+ *@Author: dingjiawen
+ *@CreateTime: 2023-01-05  10:25
+ *@Description:
+ *  TODO  二刷leecode145题  二叉树的后序遍历:
+ *      给你一棵二叉树的根节点 root ，返回其节点值的 后序遍历 。
+ *@Version: 1.0
+ */
+public class T02_PostOrderTraversal {
+
+    @Test
+    public void test() {
+        ArrayList<Integer> list = new ArrayList<>(Arrays.asList(3,1,2,5,6,7,8,10));
+        TreeNode root = TreeUtils.structureTree(list, 0);
+        System.out.println(postorderTraversal3(root));
+
+    }
+
+    /**
+     * 思路:递归法
+     * @param root
+     * @return
+     */
+    List<Integer> result;
+    public List<Integer> postorderTraversal(TreeNode root) {
+        result=new ArrayList<>();
+        postTraversal(root);
+        return result;
+
+    }
+    public void postTraversal(TreeNode root) {
+        if(root==null){
+            return;
+        }
+
+        postTraversal(root.left);
+        postTraversal(root.right);
+        result.add(root.val);
+
+    }
+
+
+    /**
+     * 思路:stack迭代法:中右左——>反序：左右中
+     * @param root
+     * @return
+     */
+    public List<Integer> postorderTraversal1(TreeNode root) {
+        List<Integer> result = new ArrayList<>();
+
+        if(root==null){
+            return result;
+        }
+
+        Stack<TreeNode> stack = new Stack<>();
+        stack.push(root);
+
+        while(!stack.isEmpty()){
+
+            TreeNode node = stack.pop();
+            result.add(node.val);
+            if(node.left!=null){
+                stack.push(node.left);
+            }
+            if(node.right!=null){
+                stack.push(node.right);
+            }
+
+        }
+
+        int size = result.size();
+        List<Integer> result1 = new ArrayList<>();
+        while (size>0){
+            result1.add(result.get(--size));
+        }
+        return result1;
+
+    }
+
+
+    /**
+     * 思路:优化stack迭代法:改变树结构
+     * @param root
+     * @return
+     */
+    public List<Integer> postorderTraversal2(TreeNode root) {
+        List<Integer> result = new ArrayList<>();
+
+        if(root==null){
+            return result;
+        }
+
+        Stack<TreeNode> stack = new Stack<>();
+        TreeNode cur=root;
+
+        while(cur!=null||!stack.isEmpty()){
+
+            if(cur!=null){
+                stack.push(cur);
+                cur=cur.left;
+            }else {
+                TreeNode node = stack.peek();
+                if(node.right!=null){
+                    cur=node.right;
+                    node.right=null;//核心是把right置空，防止下次在遍历到,但是会改变原本的树结构
+                }else {
+                    //右边没有了,安心处理node
+                    result.add(node.val);
+                    stack.pop();
+                }
+            }
+
+        }
+
+
+        return result;
+
+    }
+
+    /**
+     * 官方不改变树结构的stack法，使用一个pre记录之前的node,如果之前的node=root.right就说明右边的已经遍历完了
+     * @param root
+     * @return
+     */
+    public List<Integer> postorderTraversal4(TreeNode root) {
+        List<Integer> res = new ArrayList<Integer>();
+        if (root == null) {
+            return res;
+        }
+
+        Deque<TreeNode> stack = new LinkedList<TreeNode>();
+        TreeNode prev = null;
+        while (root != null || !stack.isEmpty()) {
+            while (root != null) {
+                stack.push(root);
+                root = root.left;
+            }
+            root = stack.pop();
+            if (root.right == null || root.right == prev) {
+                //核心在于prev指针记录之前是否遍历他的右边
+                res.add(root.val);
+                prev = root;
+                root = null;
+            } else {
+                stack.push(root);
+                root = root.right;
+            }
+        }
+        return res;
+    }
+
+
+
+
+    /**
+     * 思路:Morris遍历：在不改变结构的情况下直接的Morris遍历是不行的，还是得中右左然后反向
+     * @param root
+     * @return
+     */
+    public List<Integer> postorderTraversal3(TreeNode root) {
+        List<Integer> result = new ArrayList<>();
+
+        if(root==null){
+            return result;
+        }
+
+        TreeNode cur=root;
+
+        while (cur!=null){
+            TreeNode node=cur.right;
+
+            if(node!=null){
+                //存在右节点，寻找右子节点的最左节点
+                while (node.left!=cur&&node.left!=null){
+                    node=node.left;
+                }
+                //判断是从啥条件出来的
+                if(node.left==null){
+                    //第一次遍历到这
+                    node.left=cur;//将其连接
+                    result.add(cur.val);
+                    //放心将节点右移
+                    cur=cur.right;
+                }else if(node.left==cur){
+                    //第二次遍历到这
+                    cur=cur.left;
+                    node.left=null;
+                }
+            }else {
+                //右边为空了,遍历左节点去
+                result.add(cur.val);
+                cur=cur.left;
+            }
+        }
+
+
+        int size = result.size();
+        List<Integer> result1 = new ArrayList<>();
+        while (size>0){
+            result1.add(result.get(--size));
+        }
+        return result1;
+
+
+
+
+    }
+
+
+}

Leecode/src/main/java/com/markilue/leecode/tree/second/T03_InorderTraversal.java

@@ -0,0 +1,119 @@
+package com.markilue.leecode.tree.second;
+
+import com.markilue.leecode.tree.TreeNode;
+import com.markilue.leecode.tree.TreeUtils;
+import org.junit.Test;
+
+import java.util.ArrayList;
+import java.util.Arrays;
+import java.util.List;
+import java.util.Stack;
+
+/**
+ *@BelongsProject: Leecode
+ *@BelongsPackage: com.markilue.leecode.tree.second
+ *@Author: dingjiawen
+ *@CreateTime: 2023-01-05  11:35
+ *@Description:
+ * TODO  二刷力扣94题  二叉树的中序遍历:
+ *  给定一个二叉树的根节点 root ，返回 它的 中序 遍历 。
+ *@Version: 1.0
+ */
+public class T03_InorderTraversal {
+
+    @Test
+    public void test() {
+        ArrayList<Integer> list = new ArrayList<>(Arrays.asList(3,1,2,5,6,7,8,10));
+        TreeNode root = TreeUtils.structureTree(list, 0);
+        System.out.println(inorderTraversal2(root));
+
+    }
+
+    /**
+     * 思路:递归法
+     * @param root
+     * @return
+     */
+    List<Integer> result;
+    public List<Integer> inorderTraversal(TreeNode root) {
+        result=new ArrayList<>();
+        midTraversal(root);
+        return result;
+    }
+    public void midTraversal(TreeNode root) {
+        if(root==null){
+            return;
+        }
+        midTraversal(root.left);
+        result.add(root.val);
+        midTraversal(root.right);
+    }
+
+
+    /**
+     * 思路:stack非递归法
+     * @param root
+     * @return
+     */
+    public List<Integer> inorderTraversal1(TreeNode root) {
+        ArrayList<Integer> result = new ArrayList<>();
+        if(root==null){
+            return result;
+        }
+        Stack<TreeNode> stack = new Stack<>();
+        TreeNode cur=root;
+
+        while (cur!=null||!stack.isEmpty()){
+            if(cur!=null){
+                stack.push(cur);
+                cur=cur.left;
+            }else {
+                TreeNode node = stack.pop();
+                result.add(node.val);
+                cur=node.right;//看看他的右边
+            }
+        }
+
+        return result;
+    }
+
+
+    /**
+     * 思路:Morris遍历
+     * @param root
+     * @return
+     */
+    public List<Integer> inorderTraversal2(TreeNode root) {
+        List<Integer> result = new ArrayList<>();
+        if(root==null){
+            return result;
+        }
+
+        TreeNode cur=root;
+        while (cur!=null){
+            TreeNode node=cur.left;
+            if(node!=null){
+                while(node.right!=cur&&node.right!=null){
+                    node=node.right;
+                }
+
+                if(node.right==null){
+                    node.right=cur;
+                    cur=cur.left;
+                }else {
+                    //左边节点遍历完了
+                    node.right=null;
+                    result.add(cur.val);
+                    cur=cur.right;
+                }
+            }else {
+                result.add(cur.val);//左边节点空了
+                cur=cur.right;
+            }
+        }
+
+
+        return result;
+    }
+
+}

Leecode/src/main/java/com/markilue/leecode/tree/second/T041_LevelOrderBottom.java

@@ -0,0 +1,118 @@
+package com.markilue.leecode.tree.second;
+
+import com.markilue.leecode.tree.TreeNode;
+import com.markilue.leecode.tree.TreeUtils;
+import org.junit.Test;
+
+import java.util.*;
+
+/**
+ *@BelongsProject: Leecode
+ *@BelongsPackage: com.markilue.leecode.tree.second
+ *@Author: dingjiawen
+ *@CreateTime: 2023-01-05  12:15
+ *@Description:
+ * TODO  力扣107题  二叉树的层序遍历 II:
+ *  给你二叉树的根节点 root ，返回其节点值 自底向上的层序遍历 。
+ *  （即按从叶子节点所在层到根节点所在的层，逐层从左向右遍历）
+ *@Version: 1.0
+ */
+public class T041_LevelOrderBottom {
+
+    @Test
+    public void test() {
+        ArrayList<Integer> list = new ArrayList<>(Arrays.asList(3,9,20,null,null,15,7));
+        TreeNode root = TreeUtils.structureTree(list, 0);
+        System.out.println(levelOrderBottom1(root));
+
+    }
+
+    /**
+     * 思路：queue先层序后反向
+     * 速度击败91.28%，内存击败6.48%
+     * @param root
+     * @return
+     */
+    public List<List<Integer>> levelOrderBottom(TreeNode root) {
+        List<List<Integer>> result = new ArrayList<>();
+        if(root==null){
+            return result;
+        }
+
+        Deque<TreeNode> queue = new LinkedList<>();
+        queue.offer(root);
+
+        while (!queue.isEmpty()){
+            ArrayList<Integer> cur = new ArrayList<>();
+            int size = queue.size();
+            for (int i = 0; i < size; i++) {
+                TreeNode node = queue.poll();
+                cur.add(node.val);
+                if(node.left!=null)queue.offer(node.left);
+                if(node.right!=null)queue.offer(node.right);
+            }
+            result.add(cur);
+        }
+        Collections.reverse(result);
+        return result;
+    }
+
+
+    /**
+     * 思路：dfs深度优先递归
+     * @param root
+     * @return
+     */
+    List<List<Integer>> result = new ArrayList<>();
+    int maxDepth=0;
+    public List<List<Integer>> levelOrderBottom1(TreeNode root) {
+
+        dfs(root,0);
+        Collections.reverse(result);
+        return result;
+    }
+    public void dfs(TreeNode root,int deep) {
+        if(root==null){
+            return;
+        }
+        if(result.size()==deep){
+            result.add(new ArrayList<>());
+        }
+        dfs(root.left,deep+1);
+        dfs(root.right,deep+1);
+        result.get(deep).add(root.val);
+    }
+
+
+    /**
+     * 官方直接加在头部法，避免反向
+     * @param root
+     * @return
+     */
+    public List<List<Integer>> levelOrderBottom2(TreeNode root) {
+        List<List<Integer>> levelOrder = new LinkedList<List<Integer>>();
+        if (root == null) {
+            return levelOrder;
+        }
+        Queue<TreeNode> queue = new LinkedList<TreeNode>();
+        queue.offer(root);
+        while (!queue.isEmpty()) {
+            List<Integer> level = new ArrayList<Integer>();
+            int size = queue.size();
+            for (int i = 0; i < size; i++) {
+                TreeNode node = queue.poll();
+                level.add(node.val);
+                TreeNode left = node.left, right = node.right;
+                if (left != null) {
+                    queue.offer(left);
+                }
+                if (right != null) {
+                    queue.offer(right);
+                }
+            }
+            levelOrder.add(0, level);
+        }
+        return levelOrder;
+    }
+
+}

Leecode/src/main/java/com/markilue/leecode/tree/second/T04_LevelOrder.java

@@ -0,0 +1,78 @@
+package com.markilue.leecode.tree.second;
+
+import com.markilue.leecode.tree.TreeNode;
+import com.markilue.leecode.tree.TreeUtils;
+import org.junit.Test;
+
+import java.util.*;
+
+/**
+ *@BelongsProject: Leecode
+ *@BelongsPackage: com.markilue.leecode.tree.second
+ *@Author: dingjiawen
+ *@CreateTime: 2023-01-05  11:58
+ *@Description:
+ * TODO  二刷力扣102题  二叉树的层序遍历:
+ *  给你二叉树的根节点 root ，返回其节点值的 层序遍历 。 （即逐层地，从左到右访问所有节点）。
+ *@Version: 1.0
+ */
+public class T04_LevelOrder {
+
+    @Test
+    public void test() {
+        ArrayList<Integer> list = new ArrayList<>(Arrays.asList(3,9,20,null,null,15,7));
+        TreeNode root = TreeUtils.structureTree(list, 0);
+        System.out.println(levelOrder(root));
+
+    }
+
+    /**
+     * 思路:队列法
+     * @param root
+     * @return
+     */
+    public List<List<Integer>> levelOrder(TreeNode root) {
+        List<List<Integer>> result = new ArrayList<>();
+        if(root==null){
+            return result;
+        }
+
+        Deque<TreeNode> queue = new LinkedList<>();
+        queue.offer(root);
+
+        while (!queue.isEmpty()){
+            int size = queue.size();
+            List<Integer> cur = new ArrayList<>();//每层result
+            for (int i = 0; i < size; i++) {
+                TreeNode node = queue.poll();
+                cur.add(node.val);
+                if(node.left!=null)queue.offer(node.left);
+                if(node.right!=null)queue.offer(node.right);
+            }
+            result.add(cur);
+        }
+        return result;
+
+    }
+
+    /**
+     * 官方最快：递归法，虽然是深度优先，但是他通过ans.get(depth)获取到第deep层的数列
+     */
+    List<List<Integer>> ans = new ArrayList<>();
+    public List<List<Integer>> levelOrder2(TreeNode root) {
+        dfs(root, 0);
+        return ans;
+    }
+
+    public void dfs(TreeNode node, int depth) {
+        if (node == null) {
+            return;
+        }
+        if (ans.size() == depth) {
+            ans.add(new ArrayList<>());
+        }
+        ans.get(depth).add(node.val);
+        dfs(node.left, depth + 1);
+        dfs(node.right, depth + 1);
+    }
+}