diff --git a/Leecode/src/main/java/com/markilue/leecode/tree/T10_SumOfLeftLeaves.java b/Leecode/src/main/java/com/markilue/leecode/tree/T10_SumOfLeftLeaves.java
new file mode 100644
index 0000000..75c3cbb
--- /dev/null
+++ b/Leecode/src/main/java/com/markilue/leecode/tree/T10_SumOfLeftLeaves.java
@@ -0,0 +1,133 @@
+package com.markilue.leecode.tree;
+
+import org.junit.Test;
+
+import java.util.*;
+
+/**
+ *@BelongsProject: Leecode
+ *@BelongsPackage: com.markilue.leecode.tree.second
+ *@Author: dingjiawen
+ *@CreateTime: 2023-01-10  10:12
+ *@Description:
+ * TODO  力扣404  左子树之和:
+ *  给定二叉树的根节点 root ，返回所有左叶子之和。
+ *@Version: 1.0
+ */
+public class T10_SumOfLeftLeaves {
+
+    @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(sumOfLeftLeaves(root));
+
+    }
+
+    @Test
+    public void test1() {
+        ArrayList<Integer> list = new ArrayList<>(Arrays.asList(1, 2, 3, 4, 5));
+        TreeNode root = TreeUtils.structureTree(list, 0);
+        System.out.println(sumOfLeftLeaves(root));
+
+    }
+
+    /**
+     * 思路:左叶子节点(叶子节点：左右没有节点，且是左节点)    迭代法中序遍历不处理中节点即可
+     * 速度击败10.5%，内存击败80.37%
+     * @param root
+     * @return
+     */
+    public int sumOfLeftLeaves(TreeNode root) {
+        int sum = 0;
+        if (root == null) {
+            return sum;
+        }
+        Stack<TreeNode> stack = new Stack<>();
+        TreeNode cur = root;
+        TreeNode jin = root;//不让他加入的节点
+        while (cur != null || !stack.isEmpty()) {
+            if (cur != null) {
+                if (cur != jin && cur.left == null && cur.right == null) {
+                    sum += cur.val;
+                }
+                stack.push(cur);
+                cur = cur.left;
+            } else {
+                cur = stack.pop();
+                cur = cur.right;
+                jin = cur;//刚刚进入右节点则不能让他加
+            }
+        }
+        return sum;
+
+    }
+
+    /**
+     * 思路:左叶子节点(叶子节点：左右没有节点，且是左节点)    递归法中序遍历不处理中节点即可
+     * 速度击败100%，内存击败84.12%
+     * @param root
+     * @return
+     */
+    public int sumOfLeftLeaves1(TreeNode root) {
+        if (root == null) {
+            return 0;
+        }
+        return travelLeftLeaves1(root, null);
+    }
+
+    /**
+     * 遍历，看看node是不是pre的左节点来判断
+     * @param node
+     * @param pre
+     * @return
+     */
+    public int travelLeftLeaves1(TreeNode node, TreeNode pre) {
+        if (node == null) {//节点为null不用继续遍历
+            return 0;
+        }
+        if (pre != null && pre.left == node && node.left == null && node.right == null) {
+            return node.val;//左叶子节点,加上这个值
+        }
+        return travelLeftLeaves1(node.left, node) + travelLeftLeaves1(node.right, node);
+    }
+
+
+    /**
+     * 官方BFS:之前都是dfs这里试试BFS法，核心在于不是处理当前节点，而是在父节点判断左节点是不是叶子结点
+     * 速度击败100%，内存击败47.69%
+     * @param root
+     * @return
+     */
+    public int sumOfLeftLeaves3(TreeNode root) {
+        if (root == null) {
+            return 0;
+        }
+
+        Queue<TreeNode> queue = new LinkedList<TreeNode>();
+        queue.offer(root);
+        int ans = 0;
+        while (!queue.isEmpty()) {
+            TreeNode node = queue.poll();
+            if (node.left != null) {
+                if (isLeafNode(node.left)) {
+                    ans += node.left.val;
+                } else {
+                    queue.offer(node.left);
+                }
+            }
+            if (node.right != null) {
+                if (!isLeafNode(node.right)) {
+                    queue.offer(node.right);
+                }
+            }
+        }
+        return ans;
+    }
+
+    public boolean isLeafNode(TreeNode node) {
+        return node.left == null && node.right == null;
+    }
+
+
+}
diff --git a/Leecode/src/main/java/com/markilue/leecode/tree/T11_FindBottomLeftValue.java b/Leecode/src/main/java/com/markilue/leecode/tree/T11_FindBottomLeftValue.java
new file mode 100644
index 0000000..87e1203
--- /dev/null
+++ b/Leecode/src/main/java/com/markilue/leecode/tree/T11_FindBottomLeftValue.java
@@ -0,0 +1,75 @@
+package com.markilue.leecode.tree;
+
+import java.util.ArrayList;
+import java.util.LinkedList;
+import java.util.List;
+import java.util.Queue;
+
+/**
+ *@BelongsProject: Leecode
+ *@BelongsPackage: com.markilue.leecode.tree
+ *@Author: dingjiawen
+ *@CreateTime: 2023-01-10  10:53
+ *@Description:
+ * TODO  力扣513题  找左下角的值:
+ *  给定一个二叉树的 根节点 root，请找出该二叉树的 最底层 最左边 节点的值。
+ *  假设二叉树中至少有一个节点。
+ *@Version: 1.0
+ */
+public class T11_FindBottomLeftValue {
+
+    /**
+     * 思路:不仅要最底层，还要最左；事实上层序遍历就是一个很好地选择
+     * 速度击败13.85%，内存击败45.94%   2ms
+     * @param root
+     * @return
+     */
+    public int findBottomLeftValue(TreeNode root) {
+        int value = 0;
+        if (root == null) {
+            return value;
+        }
+        Queue<TreeNode> queue = new LinkedList<>();
+        queue.offer(root);
+        while (!queue.isEmpty()) {
+            int size = queue.size();
+            for (int i = 0; i < size; i++) {
+                TreeNode node = queue.poll();
+                if (i == 0) {
+                    //每层的最左边
+                    value = node.val;
+                }
+                if (node.left != null) queue.offer(node.left);
+                if (node.right != null) queue.offer(node.right);
+            }
+        }
+        return value;
+
+    }
+
+
+    /**
+     * 思路:不仅要最底层，还要最左；事实上层序遍历就是一个很好地选择,尝试递归DFS
+     * 速度击败100%，内存击败65.25%   0ms
+     * @param root
+     * @return
+     */
+    public int findBottomLeftValue1(TreeNode root) {
+        find(root,0);
+        return result.get(result.size()-1);
+
+    }
+
+    //记录每层最右
+    List<Integer> result=new ArrayList<>();
+    public void find(TreeNode node, int deep) {
+        if (node == null) {
+            return;
+        }
+        if(result.size()==deep){
+            result.add(node.val);
+        }
+        find(node.left,deep+1);
+        find(node.right,deep+1);
+    }
+}
diff --git a/Leecode/src/main/java/com/markilue/leecode/tree/HasPathSum.java b/Leecode/src/main/java/com/markilue/leecode/tree/T12_HasPathSum.java
similarity index 99%
rename from Leecode/src/main/java/com/markilue/leecode/tree/HasPathSum.java
rename to Leecode/src/main/java/com/markilue/leecode/tree/T12_HasPathSum.java
index 937151e..19b6420 100644
--- a/Leecode/src/main/java/com/markilue/leecode/tree/HasPathSum.java
+++ b/Leecode/src/main/java/com/markilue/leecode/tree/T12_HasPathSum.java
@@ -18,7 +18,7 @@ import java.util.Stack;
  * 叶子节点 是指没有子节点的节点。
  * @Version: 1.0
  */
-public class HasPathSum {
+public class T12_HasPathSum {
 
     @Test
     public void test() {
diff --git a/Leecode/src/main/java/com/markilue/leecode/tree/PathSum.java b/Leecode/src/main/java/com/markilue/leecode/tree/T13_PathSum.java
similarity index 99%
rename from Leecode/src/main/java/com/markilue/leecode/tree/PathSum.java
rename to Leecode/src/main/java/com/markilue/leecode/tree/T13_PathSum.java
index f58d50a..9803ef2 100644
--- a/Leecode/src/main/java/com/markilue/leecode/tree/PathSum.java
+++ b/Leecode/src/main/java/com/markilue/leecode/tree/T13_PathSum.java
@@ -14,7 +14,7 @@ import java.util.*;
  *  给你二叉树的根节点 root 和一个整数目标和 targetSum ，找出所有 从根节点到叶子节点 路径总和等于给定目标和的路径。
  * @Version: 1.0
  */
-public class PathSum {
+public class T13_PathSum {
 
 
     @Test
diff --git a/Leecode/src/main/java/com/markilue/leecode/tree/BuildTree.java b/Leecode/src/main/java/com/markilue/leecode/tree/T14_BuildTree.java
similarity index 99%
rename from Leecode/src/main/java/com/markilue/leecode/tree/BuildTree.java
rename to Leecode/src/main/java/com/markilue/leecode/tree/T14_BuildTree.java
index 8f9c5e5..1748a99 100644
--- a/Leecode/src/main/java/com/markilue/leecode/tree/BuildTree.java
+++ b/Leecode/src/main/java/com/markilue/leecode/tree/T14_BuildTree.java
@@ -13,7 +13,7 @@ import java.util.*;
  * 给定两个整数数组 inorder 和 postorder ，其中 inorder 是二叉树的中序遍历， postorder 是同一棵树的后序遍历，请你构造并返回这颗二叉树。
  * @Version: 1.0
  */
-public class BuildTree {
+public class T14_BuildTree {
 
 
     @Test
diff --git a/Leecode/src/main/java/com/markilue/leecode/tree/second/T12_HasPathSum.java b/Leecode/src/main/java/com/markilue/leecode/tree/second/T12_HasPathSum.java
new file mode 100644
index 0000000..7bef889
--- /dev/null
+++ b/Leecode/src/main/java/com/markilue/leecode/tree/second/T12_HasPathSum.java
@@ -0,0 +1,83 @@
+package com.markilue.leecode.tree.second;
+
+import com.markilue.leecode.tree.TreeNode;
+import org.junit.Test;
+
+import java.util.Stack;
+
+/**
+ *@BelongsProject: Leecode
+ *@BelongsPackage: com.markilue.leecode.tree.second
+ *@Author: dingjiawen
+ *@CreateTime: 2023-01-10  11:33
+ *@Description:
+ * TODO  力扣112题  二刷路径总和:
+ *  给你二叉树的根节点 root 和一个表示目标和的整数 targetSum 。
+ *  判断该树中是否存在 根节点到叶子节点 的路径，这条路径上所有节点值相加等于目标和 targetSum 。
+ *  如果存在，返回 true ；否则，返回 false 。
+ *  叶子节点 是指没有子节点的节点。
+ *@Version: 1.0
+ */
+public class T12_HasPathSum {
+    @Test
+    public void test(){
+        Integer sum=1000;
+        int i = sum.intValue();
+        System.out.println(i);
+    }
+
+    /**
+     * 思路:核心就在于在叶子结点时就判断targetSum;DFS
+     * @param root
+     * @param targetSum
+     * @return
+     */
+    public boolean hasPathSum(TreeNode root, int targetSum) {
+        if (root == null) {//避免开始就是空的情况
+            return false;
+        }
+        if (root.left == null && root.right == null) {//在叶子结点就处理逻辑
+            return root.val == targetSum;
+        }
+        int nextTarget = targetSum - root.val;
+        return hasPathSum(root.left, nextTarget) || hasPathSum(root.right, nextTarget);
+
+    }
+
+    /**
+     * 思路:非递归法DFS
+     * 速度击败6.83%，内存击败58.93%
+     * @param root
+     * @param targetSum
+     * @return
+     */
+    public boolean hasPathSum1(TreeNode root, int targetSum) {
+        if (root == null) {//避免开始就是空的情况
+            return false;
+        }
+        Stack<TreeNode> stack = new Stack<>();
+        Stack<Integer> stackSum = new Stack<>();
+        stack.push(root);
+        stackSum.push(0);
+        while (!stack.isEmpty()){
+            TreeNode node = stack.pop();
+            Integer sum = stackSum.pop();
+            sum+=node.val;
+            if(node.left==null&&node.right==null){
+                if(sum.intValue()==targetSum){
+                    return true;
+                }
+            }
+            if(node.left!=null){
+                stack.push(node.left);
+                stackSum.push(sum);
+            }
+            if(node.right!=null){
+                stack.push(node.right);
+                stackSum.push(sum);
+            }
+        }
+        return false;
+
+    }
+}
diff --git a/Leecode/src/main/java/com/markilue/leecode/tree/second/T13_PathSum.java b/Leecode/src/main/java/com/markilue/leecode/tree/second/T13_PathSum.java
new file mode 100644
index 0000000..d3df09a
--- /dev/null
+++ b/Leecode/src/main/java/com/markilue/leecode/tree/second/T13_PathSum.java
@@ -0,0 +1,103 @@
+package com.markilue.leecode.tree.second;
+
+import com.markilue.leecode.tree.TreeNode;
+
+import java.util.ArrayList;
+import java.util.List;
+
+/**
+ *@BelongsProject: Leecode
+ *@BelongsPackage: com.markilue.leecode.tree.second
+ *@Author: dingjiawen
+ *@CreateTime: 2023-01-10  11:52
+ *@Description:
+ * TODO  二刷力扣113题  路径总和II:
+ *  给你二叉树的根节点 root 和一个整数目标和 targetSum ，找出所有 从根节点到叶子节点 路径总和等于给定目标和的路径。
+ *  叶子节点 是指没有子节点的节点。
+ *@Version: 1.0
+ */
+public class T13_PathSum {
+
+    List<Integer> cur=new ArrayList<>();
+    List<List<Integer>> result=new ArrayList<>();
+
+    /**
+     * 递归法:
+     * 速度击败99.98%，内存击败88.17%
+     * @param root
+     * @param targetSum
+     * @return
+     */
+    public List<List<Integer>> pathSum(TreeNode root, int targetSum) {
+        travel(root,targetSum);
+        return result;
+
+    }
+    public void travel(TreeNode node, int targetSum){
+        if(node==null){
+            return;
+        }
+        cur.add(node.val);
+        if(node.left==null&&node.right==null){
+            //到达叶子节点
+            if(node.val==targetSum){
+                result.add(new ArrayList<>(cur));
+            }
+        }
+        travel(node.left,targetSum- node.val);
+        travel(node.right,targetSum- node.val);
+        cur.remove(cur.size()-1);
+    }
+
+
+    /**
+     * 官方最快：思路一样，可能快在使用数组进行增减
+     */
+    private List<List<Integer>> ans = new ArrayList<>();
+
+    private void dfs(TreeNode root, int targetSum, int current, int[] node, int index) {
+        if( root == null ) {
+            return;
+        }
+
+        node[index] = root.val;
+        current += root.val;
+        if( root.left == null && root.right == null ) {
+            if( current == targetSum ) {
+                List<Integer> result = new ArrayList<>();
+                for( int i = 0; i <= index; i++ ) {
+                    result.add(node[i]);
+                }
+                ans.add(result);
+            }
+            return;
+        }
+
+
+        dfs(root.left, targetSum, current, node, index + 1);
+        dfs(root.right, targetSum, current, node, index + 1);
+
+    }
+
+    private int getDepth(TreeNode root) {
+        if( root == null ) {
+            return 0;
+        }
+        if( root.left == null && root.right == null ) {
+            return 1;
+        }
+
+        return Math.max(getDepth(root.left), getDepth(root.right)) + 1;
+    }
+
+    public List<List<Integer>> pathSum1(TreeNode root, int targetSum) {
+        int depth = getDepth(root);
+        if( root == null ) {
+            return ans;
+        }
+
+        int[] node = new int[depth];
+        dfs(root, targetSum, 0, node, 0);
+        return ans;
+    }
+}
diff --git a/Leecode/src/main/java/com/markilue/leecode/tree/second/T14_BuildTree.java b/Leecode/src/main/java/com/markilue/leecode/tree/second/T14_BuildTree.java
new file mode 100644
index 0000000..5b9a1ef
--- /dev/null
+++ b/Leecode/src/main/java/com/markilue/leecode/tree/second/T14_BuildTree.java
@@ -0,0 +1,83 @@
+package com.markilue.leecode.tree.second;
+
+import com.markilue.leecode.tree.TreeNode;
+import org.junit.Test;
+
+import java.util.Arrays;
+
+/**
+ *@BelongsProject: Leecode
+ *@BelongsPackage: com.markilue.leecode.tree.second
+ *@Author: dingjiawen
+ *@CreateTime: 2023-01-10  12:18
+ *@Description:
+ * TODO  二刷力扣106题  从中序与后序遍历序列构造二叉树:
+ *  给定两个整数数组 inorder 和 postorder ，其中 inorder 是二叉树的中序遍历，
+ *  postorder 是同一棵树的后序遍历，请你构造并返回这颗 二叉树 。
+ *@Version: 1.0
+ */
+public class T14_BuildTree {
+
+    @Test
+    public void test(){
+        int[] inorder = {9, 3, 15, 20, 7}, postorder = {9, 15, 7, 20, 3};
+        System.out.println(buildTree1(inorder,postorder));
+    }
+
+    /**
+     * 思路：后续遍历的最后一个就是根节点，以此划分左右树，
+     * 自己被卡在就算划分了左右但是不知道inorder和postOrder应该怎么划分；
+     * 事实上两个个order一定一样长，可以根据中序定边界，从而划分两边
+     * 下属写法是最容易理解的写法，还有一些较为难理解但是传递索引而不是复制的写法来处理的写法，会比这个快
+     * 速度击败13.81%，内存击败15.72%  6ms
+     * @param inorder
+     * @param postorder
+     * @return
+     */
+    public TreeNode buildTree(int[] inorder, int[] postorder) {
+        int length = postorder.length;
+        if(length==0){
+            return null;
+        }
+        TreeNode root = new TreeNode(postorder[length - 1]);
+        int boundary=0;
+        for (; boundary < inorder.length; boundary++) {
+            if(inorder[boundary]==postorder[length-1]){
+                break;
+            }
+        }
+        int[] leftInorder = Arrays.copyOfRange(inorder, 0, boundary);
+        int[] rightInorder = Arrays.copyOfRange(inorder, boundary+1, length);
+        int[] leftPostorder = Arrays.copyOfRange(postorder, 0, boundary);
+        int[] rightPostorder = Arrays.copyOfRange(postorder, boundary, length - 1);
+        root.left=buildTree(leftInorder,leftPostorder);
+        root.right=buildTree(rightInorder,rightPostorder);
+
+        return root;
+
+    }
+
+
+    /**
+     * 官方最快
+     * 速度击败100%，内存击败72.96%
+     */
+    int post,in;
+    public TreeNode buildTree1(int[] inorder, int[] postorder) {
+        post=inorder.length-1;
+        in=inorder.length-1;
+        return buildTreeHelper(inorder,postorder,Integer.MAX_VALUE);
+    }
+    public TreeNode buildTreeHelper(int[] inorder, int[] postorder, int stop) {
+        if(post==-1)return null;
+        if(inorder[in]==stop){//stop是right边界，如果触及了右边界就那边就没有值了
+            in--;
+            return null;//找到了右边的边界
+        }
+        int root_val=postorder[post--];
+        TreeNode root = new TreeNode(root_val);
+        root.right=buildTreeHelper(inorder,postorder,root_val);
+        root.left=buildTreeHelper(inorder,postorder,stop);
+        return root;
+    }
+}