diff --git a/Hard/124 Binary Tree Maximum Path Sum.md b/Hard/124 Binary Tree Maximum Path Sum.md
new file mode 100644
index 0000000..6bac9d9
--- /dev/null
+++ b/Hard/124 Binary Tree Maximum Path Sum.md	
@@ -0,0 +1,60 @@
+# 124. Binary Tree Maximum Path Sum
+
+## Intuition
+
+The maximum path sum in a binary tree can be found by considering all possible paths that pass through each node. For any given node, the maximum path sum could be:
+
+1. The node's value itself
+2. The node's value plus the maximum path from its left subtree
+3. The node's value plus the maximum path from its right subtree
+4. The node's value plus both left and right maximum paths
+
+## Approach
+
+We use a post-order traversal approach to solve this problem:
+
+1. For each node, we recursively calculate the maximum path sum from its left and right subtrees
+2. We update the global maximum by considering all possible combinations:
+   - The current node's value alone
+   - Current node + left path
+   - Current node + right path
+   - Current node + left path + right path
+3. We return the maximum path sum that can be extended from the current node to its parent (which can only include at most one of its children)
+
+## Complexity
+
+- Time complexity: O(n)
+- Space complexity: O(h)
+
+## Keywords
+
+- Binary Tree
+- Post-order Traversal
+- Recursion
+
+## Code
+
+```go
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func maxPathSum(root *TreeNode) int {
+    ret := math.MinInt
+    var postOrder func(node *TreeNode) int
+    postOrder = func(node *TreeNode) int {
+        if node == nil {
+            return 0
+        }
+        left, right := postOrder(node.Left), postOrder(node.Right)
+        ret = max(ret, left + right + node.Val, left + node.Val, right + node.Val, node.Val)
+        return max(left + node.Val, right + node.Val, node.Val)
+    }
+    ret = max(ret, postOrder(root))
+    return ret
+}
+```
diff --git a/Hard/834 Sum of Distances in Tree.md b/Hard/834 Sum of Distances in Tree.md
new file mode 100644
index 0000000..c4cef75
--- /dev/null
+++ b/Hard/834 Sum of Distances in Tree.md	
@@ -0,0 +1,72 @@
+# 834. Sum of Distances in Tree
+
+## Intuition
+
+A key observation is that we can leverage the tree structure and parent-child relationships to compute these sums efficiently. Instead of calculating distances from scratch for each node, we can use the results from a parent node to derive the results for its children.
+
+## Approach
+
+The solution uses a two-pass DFS approach:
+
+1. First DFS (dfs1):
+   - Calculates the number of child nodes for each node
+   - Computes the initial sum of distances for the root node (0)
+   - This pass builds the foundation for the second pass
+2. Second DFS (dfs2):
+   - Uses the results from the first pass to compute distances for all other nodes
+   - For each child node, the sum of distances can be derived from its parent's sum using the formula:
+     `ret[child] = ret[parent] + (n - 2 * childNodes[child])`
+   - This formula works because moving from parent to child:
+     - Increases distance by 1 for all nodes not in the child's subtree
+     - Decreases distance by 1 for all nodes in the child's subtree
+
+## Complexity
+
+- Time complexity: O(n)
+- Space complexity: O(n)
+
+## Keywords
+
+- DFS
+- Distance Calculation
+
+## Code
+
+```go
+func sumOfDistancesInTree(n int, edges [][]int) []int {
+    ret, childNodes, tree := make([]int, n), make([]int, n), make([][]int, n)
+    for _, edge := range edges {
+        a, b := edge[0], edge[1]
+        tree[a], tree[b] = append(tree[a], b), append(tree[b], a)
+    }
+
+    var dfs1 func(cur, prev int)
+    dfs1 = func(cur, prev int) {
+        childNodes[cur] = 1
+        for _, neighbor := range tree[cur] {
+            if neighbor == prev {
+                continue
+            }
+            dfs1(neighbor, cur)
+            childNodes[cur] += childNodes[neighbor]
+            ret[cur] += ret[neighbor] + childNodes[neighbor]
+        }
+    }
+
+    var dfs2 func(cur, prev int)
+    dfs2 = func(cur, prev int) {
+        for _, neighbor := range tree[cur] {
+            if neighbor == prev {
+                continue
+            }
+            ret[neighbor] = ret[cur] + (n- 2 * childNodes[neighbor])
+            dfs2(neighbor, cur)
+        }
+    }
+
+    dfs1(0, -1)
+    dfs2(0, -1)
+
+    return ret
+}
+```
diff --git a/Medium/236 Lowest Common Ancestor of a Binary Tree.md b/Medium/236 Lowest Common Ancestor of a Binary Tree.md
new file mode 100644
index 0000000..057b591
--- /dev/null
+++ b/Medium/236 Lowest Common Ancestor of a Binary Tree.md	
@@ -0,0 +1,66 @@
+# 236. Lowest Common Ancestor of a Binary Tree
+
+## Intuition
+
+ We can use a post-order traversal approach where we first check the left and right subtrees, then process the current node.
+
+## Approach
+
+1. We use a recursive helper function `find` that takes the current node and the two target nodes p and q.
+2. The base cases are:
+   - If the current node is nil, return nil
+   - If the current node is either p or q, return the current node
+3. We recursively search in both left and right subtrees
+4. The logic for determining the LCA is:
+   - If both left and right subtrees return nil, return nil (neither p nor q found)
+   - If only one subtree returns a non-nil value, return that value (one of p or q found)
+   - If both subtrees return non-nil values, the current node is the LCA
+
+## Complexity
+
+- Time complexity: O(n)
+- Space complexity: O(h)
+
+## Keywords
+
+- Binary Tree
+- Recursion
+- DFS
+- Post-order Traversal
+- Lowest Common Ancestor
+
+## Code
+
+```go
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func find(node, p, q *TreeNode) *TreeNode {
+    if node == nil {
+        return nil
+    }
+    if node == p || node == q {
+        return node
+    }
+    left := find(node.Left, p, q)
+    right := find(node.Right, p, q)
+    if left == nil && right == nil {
+        return nil
+    }
+    if left != nil && right == nil {
+        return left
+    }
+    if left == nil && right != nil {
+        return right
+    }
+    return node
+}
+func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
+    return find(root, p, q)
+}
+```
diff --git a/README.md b/README.md
index 23c8040..fa192e2 100644
--- a/README.md
+++ b/README.md
@@ -18,9 +18,11 @@
 | 50 | Pow(x, n) | [go](/Medium/50%20Pow(x,%20n).md) | M |
 | 51 | N-Queens | [go](/Hard/51%20N-Queens.md) | H |
 | 105 | Construct Binary Tree from Preorder and Inorder Tranversal | [go](/Medium/105%20Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Tranversal.md) | M |
+| 124 | Binary Tree Maximum Path Sum | [go](Hard/124%20Binary%20Tree%20Maximum%20Path%20Sum.md) | H |
 | 135 | Candy | [go](/Hard/135%20Candy.md) | H |
 | 140 | Word Break II | [go](/Hard/140%20Word%20Break%20II.md) | H |
 | 146 | LRU Cache | [go](/Medium/146%20LRU%20Cache.md) | M |
+| 236 | Lowest Common Ancestor of a Binary Tree | [go](/Medium/236%20Lowest%20Common%20Ancestor%20of%20a%20Binary%20Tree.md) | M |
 | 239 | Sliding Window Maximum | [go](/Hard/239%20Sliding%20Window%20Maximum.md) | H |
 | 322 | Coin Change | [go](/Medium/322%20Coin%20Change.md) | M |
 | 460 | LFU Cache | [go](/Hard/460%20LFU%20Cache.md) | H |
@@ -29,6 +31,7 @@
 | 757 | Set Intersection Size At Least Two | [go](/Hard/757%20Set%20Intersection%20Size%20At%20Least%20Two.md) | H |
 | 815 | Bus Routes | [go](/Hard/815%20Bus%20Routes.md) | H |
 | 765 | Couples Holding Hands | [go](/Hard/765%20Couples%20Holding%20Hands.md) | H |
+| 834 | Sum of Distances in Tree | [go](/Hard/834%20Sum%20of%20Distances%20in%20Tree.md) | H |
 | 840 | Magic Squares In Grid | [go](/Medium/840%20Magic%20Squares%20In%20Grid.md) | M |
 | 881 | Boats to Save People | [go](/Medium/881%20Boats%20to%20Save%20People.md) | M |
 | 912 | Sort an Array | [go](/Medium/912%20Sort%20an%20Array.md) | M |
diff --git a/Week8/124 AC.png b/Week8/124 AC.png
new file mode 100644
index 0000000..376a3b1
Binary files /dev/null and b/Week8/124 AC.png differ
diff --git a/Week8/124 Binary Tree Maximum Path Sum.md b/Week8/124 Binary Tree Maximum Path Sum.md
new file mode 100644
index 0000000..6bac9d9
--- /dev/null
+++ b/Week8/124 Binary Tree Maximum Path Sum.md	
@@ -0,0 +1,60 @@
+# 124. Binary Tree Maximum Path Sum
+
+## Intuition
+
+The maximum path sum in a binary tree can be found by considering all possible paths that pass through each node. For any given node, the maximum path sum could be:
+
+1. The node's value itself
+2. The node's value plus the maximum path from its left subtree
+3. The node's value plus the maximum path from its right subtree
+4. The node's value plus both left and right maximum paths
+
+## Approach
+
+We use a post-order traversal approach to solve this problem:
+
+1. For each node, we recursively calculate the maximum path sum from its left and right subtrees
+2. We update the global maximum by considering all possible combinations:
+   - The current node's value alone
+   - Current node + left path
+   - Current node + right path
+   - Current node + left path + right path
+3. We return the maximum path sum that can be extended from the current node to its parent (which can only include at most one of its children)
+
+## Complexity
+
+- Time complexity: O(n)
+- Space complexity: O(h)
+
+## Keywords
+
+- Binary Tree
+- Post-order Traversal
+- Recursion
+
+## Code
+
+```go
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func maxPathSum(root *TreeNode) int {
+    ret := math.MinInt
+    var postOrder func(node *TreeNode) int
+    postOrder = func(node *TreeNode) int {
+        if node == nil {
+            return 0
+        }
+        left, right := postOrder(node.Left), postOrder(node.Right)
+        ret = max(ret, left + right + node.Val, left + node.Val, right + node.Val, node.Val)
+        return max(left + node.Val, right + node.Val, node.Val)
+    }
+    ret = max(ret, postOrder(root))
+    return ret
+}
+```
diff --git a/Week8/236 AC.png b/Week8/236 AC.png
new file mode 100644
index 0000000..439cbe1
Binary files /dev/null and b/Week8/236 AC.png differ
diff --git a/Week8/236 Lowest Common Ancestor of a Binary Tree.md b/Week8/236 Lowest Common Ancestor of a Binary Tree.md
new file mode 100644
index 0000000..057b591
--- /dev/null
+++ b/Week8/236 Lowest Common Ancestor of a Binary Tree.md	
@@ -0,0 +1,66 @@
+# 236. Lowest Common Ancestor of a Binary Tree
+
+## Intuition
+
+ We can use a post-order traversal approach where we first check the left and right subtrees, then process the current node.
+
+## Approach
+
+1. We use a recursive helper function `find` that takes the current node and the two target nodes p and q.
+2. The base cases are:
+   - If the current node is nil, return nil
+   - If the current node is either p or q, return the current node
+3. We recursively search in both left and right subtrees
+4. The logic for determining the LCA is:
+   - If both left and right subtrees return nil, return nil (neither p nor q found)
+   - If only one subtree returns a non-nil value, return that value (one of p or q found)
+   - If both subtrees return non-nil values, the current node is the LCA
+
+## Complexity
+
+- Time complexity: O(n)
+- Space complexity: O(h)
+
+## Keywords
+
+- Binary Tree
+- Recursion
+- DFS
+- Post-order Traversal
+- Lowest Common Ancestor
+
+## Code
+
+```go
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func find(node, p, q *TreeNode) *TreeNode {
+    if node == nil {
+        return nil
+    }
+    if node == p || node == q {
+        return node
+    }
+    left := find(node.Left, p, q)
+    right := find(node.Right, p, q)
+    if left == nil && right == nil {
+        return nil
+    }
+    if left != nil && right == nil {
+        return left
+    }
+    if left == nil && right != nil {
+        return right
+    }
+    return node
+}
+func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
+    return find(root, p, q)
+}
+```
diff --git a/Week8/834 AC.png b/Week8/834 AC.png
new file mode 100644
index 0000000..afc4021
Binary files /dev/null and b/Week8/834 AC.png differ
diff --git a/Week8/834 Sum of Distances in Tree.md b/Week8/834 Sum of Distances in Tree.md
new file mode 100644
index 0000000..c4cef75
--- /dev/null
+++ b/Week8/834 Sum of Distances in Tree.md	
@@ -0,0 +1,72 @@
+# 834. Sum of Distances in Tree
+
+## Intuition
+
+A key observation is that we can leverage the tree structure and parent-child relationships to compute these sums efficiently. Instead of calculating distances from scratch for each node, we can use the results from a parent node to derive the results for its children.
+
+## Approach
+
+The solution uses a two-pass DFS approach:
+
+1. First DFS (dfs1):
+   - Calculates the number of child nodes for each node
+   - Computes the initial sum of distances for the root node (0)
+   - This pass builds the foundation for the second pass
+2. Second DFS (dfs2):
+   - Uses the results from the first pass to compute distances for all other nodes
+   - For each child node, the sum of distances can be derived from its parent's sum using the formula:
+     `ret[child] = ret[parent] + (n - 2 * childNodes[child])`
+   - This formula works because moving from parent to child:
+     - Increases distance by 1 for all nodes not in the child's subtree
+     - Decreases distance by 1 for all nodes in the child's subtree
+
+## Complexity
+
+- Time complexity: O(n)
+- Space complexity: O(n)
+
+## Keywords
+
+- DFS
+- Distance Calculation
+
+## Code
+
+```go
+func sumOfDistancesInTree(n int, edges [][]int) []int {
+    ret, childNodes, tree := make([]int, n), make([]int, n), make([][]int, n)
+    for _, edge := range edges {
+        a, b := edge[0], edge[1]
+        tree[a], tree[b] = append(tree[a], b), append(tree[b], a)
+    }
+
+    var dfs1 func(cur, prev int)
+    dfs1 = func(cur, prev int) {
+        childNodes[cur] = 1
+        for _, neighbor := range tree[cur] {
+            if neighbor == prev {
+                continue
+            }
+            dfs1(neighbor, cur)
+            childNodes[cur] += childNodes[neighbor]
+            ret[cur] += ret[neighbor] + childNodes[neighbor]
+        }
+    }
+
+    var dfs2 func(cur, prev int)
+    dfs2 = func(cur, prev int) {
+        for _, neighbor := range tree[cur] {
+            if neighbor == prev {
+                continue
+            }
+            ret[neighbor] = ret[cur] + (n- 2 * childNodes[neighbor])
+            dfs2(neighbor, cur)
+        }
+    }
+
+    dfs1(0, -1)
+    dfs2(0, -1)
+
+    return ret
+}
+```