feat: add golang solution for lcof problem 33-35

Author: hzliangbin

lcof/面试题33. 二叉搜索树的后序遍历序列/README.md

@@ -110,7 +110,43 @@ var verifyPostorder = function(postorder) {
 };
 ```
 
+### Go
+
+```go
+func verifyPostorder(postorder []int) bool {
+    if len(postorder) < 2 {
+        return true
+    }
+    return helper(postorder, 0, len(postorder)-1)
+}
+//递归
+func helper(postorder []int , left,right int) bool {
+    if left >= right {
+        return true
+    }
+    //最后一位即根
+    rootValue := postorder[right]
+    //从左开始往右遍历，直到大于根停止,小于部分是左子树
+    i := left
+    for i < right && postorder[i] < rootValue {
+        i++
+    }
+    //剩下部分是右子树，检查是否都大于根值
+    for j := i; j < right; j++ {
+        if postorder[j] < rootValue {
+            return false
+        }
+    }
+    l := helper(postorder,left,i-1) //检查左子树，左子树i要减一
+    r := helper(postorder,i,right-1)//检查右子树，剔除最后一位是根
+    return l && r
+}
+```
+
+
+
 ### ...
+
 ```
 
 ```

lcof/面试题33. 二叉搜索树的后序遍历序列/Solution.go

@@ -0,0 +1,28 @@
+func verifyPostorder(postorder []int) bool {
+    if len(postorder) < 2 {
+        return true
+    }
+    return helper(postorder, 0, len(postorder)-1)
+}
+//递归
+func helper(postorder []int , left,right int) bool {
+    if left >= right {
+        return true
+    }
+    //最后一位即根
+    rootValue := postorder[right]
+    //从左开始往右遍历，直到大于根停止,小于部分是左子树
+    i := left
+    for i < right && postorder[i] < rootValue {
+        i++
+    }
+    //剩下部分是右子树，检查是否都大于根值
+    for j := i; j < right; j++ {
+        if postorder[j] < rootValue {
+            return false
+        }
+    }
+    l := helper(postorder,left,i-1) //检查左子树，左子树i要减一
+    r := helper(postorder,i,right-1)//检查右子树，剔除最后一位是根
+    return l && r
+}

lcof/面试题34. 二叉树中和为某一值的路径/README.md

@@ -139,7 +139,40 @@ var pathSum = function(root, sum) {
 };
 ```
 
+### Go
+
+```go
+var res [][]int 
+func pathSum(root *TreeNode, sum int) [][]int {
+    res = [][]int{}
+    if root == nil {
+        return res
+    }
+    helper(root, sum, []int{})
+    return res
+}
+
+func helper(node *TreeNode, target int, ans []int) {
+    if node == nil {
+        return
+    }
+    ans = append(ans,node.Val)
+    target -= node.Val
+    if target == 0 && node.Left == nil && node.Right == nil {
+        tmp := make([]int,len(ans))
+        copy(tmp,ans)
+        res = append(res,tmp)
+    } else {
+        helper(node.Left, target, ans)
+        helper(node.Right, target, ans)
+    }
+}
+```
+
+
+
 ### ...
+
 ```
 
 ```

lcof/面试题34. 二叉树中和为某一值的路径/Solution.go

@@ -0,0 +1,25 @@
+var res [][]int 
+func pathSum(root *TreeNode, sum int) [][]int {
+    res = [][]int{}
+    if root == nil {
+        return res
+    }
+    helper(root, sum, []int{})
+    return res
+}
+
+func helper(node *TreeNode, target int, ans []int) {
+    if node == nil {
+        return
+    }
+    ans = append(ans,node.Val)
+    target -= node.Val
+    if target == 0 && node.Left == nil && node.Right == nil {
+        tmp := make([]int,len(ans))
+        copy(tmp,ans)
+        res = append(res,tmp)
+    } else {
+        helper(node.Left, target, ans)
+        helper(node.Right, target, ans)
+    }
+}

lcof/面试题35. 复杂链表的复制/README.md

@@ -155,7 +155,58 @@ var copyRandomList = function(head) {
 };
 ```
 
+### Go
+
+```go
+func copyRandomList(head *Node) *Node {
+    if head == nil {
+        return nil
+    }
+    cur, next := head,head
+    //完成对当前节点的复制
+    for cur != nil {
+        next = cur.Next
+        cur.Next = &Node{
+            Val: cur.Val,
+            Next: next,
+            Random: nil,
+        }
+        cur = next
+    }
+    //设置复制节点的random节点
+    cur = head
+    curCopy := head
+    for cur != nil {
+        next = cur.Next.Next
+        curCopy = cur.Next
+        if cur.Random == nil {
+            curCopy.Random = nil
+        } else {
+            curCopy.Random = cur.Random.Next
+        }
+        cur = next
+    }
+    res := head.Next
+    cur = head
+    for cur != nil {
+        next = cur.Next.Next
+        curCopy = cur.Next
+        cur.Next = next
+        if next != nil {
+            curCopy.Next = next.Next
+        } else {
+            curCopy.Next = nil
+        }
+        cur = cur.Next
+    }
+    return res
+}
+```
+
+
+
 ### ...
+
 ```
 
 ```

lcof/面试题35. 复杂链表的复制/Solution.go

@@ -0,0 +1,44 @@
+func copyRandomList(head *Node) *Node {
+    if head == nil {
+        return nil
+    }
+    cur, next := head,head
+    //完成对当前节点的复制
+    for cur != nil {
+        next = cur.Next
+        cur.Next = &Node{
+            Val: cur.Val,
+            Next: next,
+            Random: nil,
+        }
+        cur = next
+    }
+    //设置复制节点的random节点
+    cur = head
+    curCopy := head
+    for cur != nil {
+        next = cur.Next.Next
+        curCopy = cur.Next
+        if cur.Random == nil {
+            curCopy.Random = nil
+        } else {
+            curCopy.Random = cur.Random.Next
+        }
+        cur = next
+    }
+    res := head.Next
+    cur = head
+    for cur != nil {
+        next = cur.Next.Next
+        curCopy = cur.Next
+        cur.Next = next
+        if next != nil {
+            curCopy.Next = next.Next
+        } else {
+            curCopy.Next = nil
+        }
+        cur = cur.Next
+    }
+    return res
+
+}