提供JavaScript版本的《二叉搜索树中的插入操作》

kok-s0s · kok-s0s · commit ad1faf8a150a · 2021-06-25T22:54:40.000+08:00
diff --git a/problems/0701.二叉搜索树中的插入操作.md b/problems/0701.二叉搜索树中的插入操作.md
@@ -320,7 +320,84 @@ var insertIntoBST = function (root, val) {
 };
 ```
 
+> 无返回值的递归
 
+```javascript
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val, left, right) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.left = (left===undefined ? null : left)
+ *     this.right = (right===undefined ? null : right)
+ * }
+ */
+/**
+ * @param {TreeNode} root
+ * @param {number} val
+ * @return {TreeNode}
+ */
+var insertIntoBST = function (root, val) {
+    let parent = new TreeNode(0);
+    const preOrder = (cur, val) => {
+        if (cur === null) {
+            let node = new TreeNode(val);
+            if (parent.val > val)
+                parent.left = node;
+            else
+                parent.right = node;
+            return;
+        }
+        parent = cur;
+        if (cur.val > val)
+            preOrder(cur.left, val);
+        if (cur.val < val)
+            preOrder(cur.right, val);
+    }
+    if (root === null)
+        root = new TreeNode(val);
+    preOrder(root, val);
+    return root;
+};
+```
+
+> 迭代
+
+```javascript
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val, left, right) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.left = (left===undefined ? null : left)
+ *     this.right = (right===undefined ? null : right)
+ * }
+ */
+/**
+ * @param {TreeNode} root
+ * @param {number} val
+ * @return {TreeNode}
+ */
+var insertIntoBST = function (root, val) {
+    if (root === null) {
+        root = new TreeNode(val);
+    } else {
+        let parent = new TreeNode(0);
+        let cur = root;
+        while (cur) {
+            parent = cur;
+            if (cur.val > val)
+                cur = cur.left;
+            else
+                cur = cur.right;
+        }
+        let node = new TreeNode(val);
+        if (parent.val > val)
+            parent.left = node;
+        else
+            parent.right = node;
+    }
+    return root;
+};
+```
 
 -----------------------
 * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
