feat: add solutions to lc problem: No.1305

No.1395.All Elements in Two Binary Search Trees

Author: yanglbme

solution/1300-1399/1305.All Elements in Two Binary Search Trees/README.md

@@ -59,22 +59,207 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+二叉树中序遍历 + 有序列表归并。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+class Solution:
+    def getAllElements(self, root1: TreeNode, root2: TreeNode) -> List[int]:
+        def dfs(root, t):
+            if root is None:
+                return
+            dfs(root.left, t)
+            t.append(root.val)
+            dfs(root.right, t)
+
+        def merge(t1, t2):
+            ans = []
+            i = j = 0
+            while i < len(t1) and j < len(t2):
+                if t1[i] <= t2[j]:
+                    ans.append(t1[i])
+                    i += 1
+                else:
+                    ans.append(t2[j])
+                    j += 1
+            while i < len(t1):
+                ans.append(t1[i])
+                i += 1
+            while j < len(t2):
+                ans.append(t2[j])
+                j += 1
+            return ans
+
+        t1, t2 = [], []
+        dfs(root1, t1)
+        dfs(root2, t2)
+        return merge(t1, t2)
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode() {}
+ *     TreeNode(int val) { this.val = val; }
+ *     TreeNode(int val, TreeNode left, TreeNode right) {
+ *         this.val = val;
+ *         this.left = left;
+ *         this.right = right;
+ *     }
+ * }
+ */
+class Solution {
+    public List<Integer> getAllElements(TreeNode root1, TreeNode root2) {
+        List<Integer> t1 = new ArrayList<>();
+        List<Integer> t2 = new ArrayList<>();
+        dfs(root1, t1);
+        dfs(root2, t2);
+        return merge(t1, t2);
+    }
+
+    private void dfs(TreeNode root, List<Integer> t) {
+        if (root == null) {
+            return;
+        }
+        dfs(root.left, t);
+        t.add(root.val);
+        dfs(root.right, t);
+    }
+
+    private List<Integer> merge(List<Integer> t1, List<Integer> t2) {
+        List<Integer> ans = new ArrayList<>();
+        int i = 0, j = 0;
+        while (i < t1.size() && j < t2.size()) {
+            if (t1.get(i) <= t2.get(j)) {
+                ans.add(t1.get(i++));
+            } else {
+                ans.add(t2.get(j++));
+            }
+        }
+        while (i < t1.size()) {
+            ans.add(t1.get(i++));
+        }
+        while (j < t2.size()) {
+            ans.add(t2.get(j++));
+        }
+        return ans;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
+ * };
+ */
+class Solution {
+public:
+    vector<int> getAllElements(TreeNode* root1, TreeNode* root2) {
+        vector<int> t1;
+        vector<int> t2;
+        dfs(root1, t1);
+        dfs(root2, t2);
+        return merge(t1, t2);
+    }
+
+    void dfs(TreeNode* root, vector<int>& t) {
+        if (!root) return;
+        dfs(root->left, t);
+        t.push_back(root->val);
+        dfs(root->right, t);
+    }
+
+    vector<int> merge(vector<int>& t1, vector<int>& t2) {
+        vector<int> ans;
+        int i = 0, j = 0;
+        while (i < t1.size() && j < t2.size())
+        {
+            if (t1[i] <= t2[j]) ans.push_back(t1[i++]);
+            else ans.push_back(t2[j++]);
+        }
+        while (i < t1.size()) ans.push_back(t1[i++]);
+        while (j < t2.size()) ans.push_back(t2[j++]);
+        return ans;
+    }
+};
+```
 
+### **Go**
+
+```go
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func getAllElements(root1 *TreeNode, root2 *TreeNode) []int {
+	var dfs func(root *TreeNode) []int
+	dfs = func(root *TreeNode) []int {
+		if root == nil {
+			return []int{}
+		}
+		left := dfs(root.Left)
+		right := dfs(root.Right)
+		left = append(left, root.Val)
+		left = append(left, right...)
+		return left
+	}
+	merge := func(t1, t2 []int) []int {
+		var ans []int
+		i, j := 0, 0
+		for i < len(t1) && j < len(t2) {
+			if t1[i] <= t2[j] {
+				ans = append(ans, t1[i])
+				i++
+			} else {
+				ans = append(ans, t2[j])
+				j++
+			}
+		}
+		for i < len(t1) {
+			ans = append(ans, t1[i])
+			i++
+		}
+		for j < len(t2) {
+			ans = append(ans, t2[j])
+			j++
+		}
+		return ans
+	}
+	t1, t2 := dfs(root1), dfs(root2)
+	return merge(t1, t2)
+}
 ```
 
 ### **...**