修改java版本解法的位置，适合图的解法放在第一个，第二个解法（难理解版本）放在其后，新增java版本的迭代法

problems/0450.删除二叉搜索树中的节点.md

@@ -268,6 +268,34 @@ public:
 
 
 ## Java 
+```java
+// 解法1(最好理解的版本)
+class Solution {
+  public TreeNode deleteNode(TreeNode root, int key) {
+    if (root == null) return root;
+    if (root.val == key) {
+      if (root.left == null) {
+        return root.right;
+      } else if (root.right == null) {
+        return root.left;
+      } else {
+        TreeNode cur = root.right;
+        while (cur.left != null) {
+          cur = cur.left;
+        }
+        cur.left = root.left;
+        root = root.right;
+        return root;
+      }
+    }
+    if (root.val > key) root.left = deleteNode(root.left, key);
+    if (root.val < key) root.right = deleteNode(root.right, key);
+    return root;
+  }
+}
+```
+
+
 ```java
 class Solution {
     public TreeNode deleteNode(TreeNode root, int key) {
@@ -296,33 +324,57 @@ class Solution {
     }
 }
 ```
+递归法
 ```java
-// 解法2
 class Solution {
   public TreeNode deleteNode(TreeNode root, int key) {
-    if (root == null) return root;
-    if (root.val == key) {
-      if (root.left == null) {
-        return root.right;
-      } else if (root.right == null) {
-        return root.left;
-      } else {
-        TreeNode cur = root.right;
-        while (cur.left != null) {
-          cur = cur.left;
-        }
-        cur.left = root.left;
-        root = root.right;
-        return root;
+    if (root == null){
+      return null;
+    }
+    //寻找对应的对应的前面的节点，以及他的前一个节点
+    TreeNode cur = root;
+    TreeNode pre = null;
+    while (cur != null){
+      if (cur.val < key){
+        pre = cur;
+        cur = cur.right;
+      } else if (cur.val > key) {
+        pre = cur;
+        cur = cur.left;
+      }else {
+        break;
       }
     }
-    if (root.val > key) root.left = deleteNode(root.left, key);
-    if (root.val < key) root.right = deleteNode(root.right, key);
+    if (pre == null){
+      return deleteOneNode(cur);
+    }
+    if (pre.left !=null && pre.left.val == key){
+      pre.left = deleteOneNode(cur);
+    }
+    if (pre.right !=null && pre.right.val == key){
+      pre.right = deleteOneNode(cur);
+    }
     return root;
   }
+
+  public TreeNode deleteOneNode(TreeNode node){
+    if (node == null){
+      return null;
+    }
+    if (node.right == null){
+      return node.left;
+    }
+    TreeNode cur = node.right;
+    while (cur.left !=null){
+      cur = cur.left;
+    }
+    cur.left = node.left;
+    return node.right;
+  }
 }
 ```
 
+
 ## Python 
 递归法（版本一）
 ```python