diff --git a/content/posts/leetcode.md b/content/posts/leetcode.md
index ff2a057..9ea00b1 100644
--- a/content/posts/leetcode.md
+++ b/content/posts/leetcode.md
@@ -5801,3 +5801,127 @@ func validpath(grid [][]int,rowlen int,mid int)bool{
     return false
 }
 ```
+
+## day 79 2024-05-16
+
+### 2331. Evaluate Boolean Binary Tree
+
+You are given the root of a full binary tree with the following properties:
+
+Leaf nodes have either the value 0 or 1, where 0 represents False and 1 represents True.
+Non-leaf nodes have either the value 2 or 3, where 2 represents the boolean OR and 3 represents the boolean AND.
+The evaluation of a node is as follows:
+
+If the node is a leaf node, the evaluation is the value of the node, i.e. True or False.
+Otherwise, evaluate the node's two children and apply the boolean operation of its value with the children's evaluations.
+Return the boolean result of evaluating the root node.
+
+A full binary tree is a binary tree where each node has either 0 or 2 children.
+
+A leaf node is a node that has zero children.
+
+![0516LXOTbf2QVjTC](https://testingcf.jsdelivr.net/gh/game-loader/picbase@master/uPic/0516LXOTbf2QVjTC.png)
+
+### 题解
+
+后续遍历计算各个节点的布尔值最终得到根节点的布尔值即可, 考查二叉树的后序遍历, 可以使用递归实现.
+
+### 代码
+
+```go
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func evaluateTree(root *TreeNode) bool {
+    result := caculatebool(root)
+    return result
+}
+
+func caculatebool(father *TreeNode)bool{
+    if father.Left == nil{
+        return father.Val == 1
+    }else {
+        if father.Val == 2{
+            return caculatebool(father.Left) || caculatebool(father.Right)
+        }else if father.Val == 3{
+            return caculatebool(father.Left) && caculatebool(father.Right)
+        }
+    }
+    return false
+}
+```
+
+## day80 2024-05-17
+
+### 1325. Delete Leaves With a Given Value
+
+Given a binary tree root and an integer target, delete all the leaf nodes with value target.
+
+Note that once you delete a leaf node with value target, if its parent node becomes a leaf node and has the value target, it should also be deleted (you need to continue doing that until you cannot).
+
+![0517CoXoi4xFlfTN](https://testingcf.jsdelivr.net/gh/game-loader/picbase@master/uPic/0517CoXoi4xFlfTN.png)
+
+### 题解
+
+后序遍历即可, 使用递归遍历过程中返回当前节点是否被删除, 需要被删除返回true, 否则返回false. 父节点对子节点进行函数调用, 根据子节点的情况决定是否判断父节点是否要被删除. 这里相当于将子节点及子节点以下的信息向上传递.
+
+### 代码
+
+```go
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func removeLeafNodes(root *TreeNode, target int) *TreeNode {
+    if root == nil{
+        return nil
+    }
+    result := deleteaNode(root, target)
+    if result{
+        return nil
+    }else{
+        return root
+    }
+}
+
+func deleteaNode(father *TreeNode, target int)bool{
+    deleted := true
+    judge := false
+    if father.Left != nil {
+        judge = deleteaNode(father.Left, target)
+        if judge{
+            father.Left = nil
+        }
+        deleted = deleted && judge
+
+    }
+    if father.Right != nil{
+        judge = deleteaNode(father.Right, target)
+        if judge{
+            father.Right = nil
+        }
+        deleted = deleted && deleteaNode(father.Right, target)
+    }
+    if deleted{
+        if father.Val == target{
+            return true
+        }
+    }
+
+    return false
+}
+
+```
+
+### 总结
+
+直接返回节点会更简洁一些, 对于值符合的叶子节点, 递归时返回nil即可.