From 55cbb97ab6983185ebe00c1387333501c6eb70a7 Mon Sep 17 00:00:00 2001
From: Joshua <47053655+Joshua-Lu@users.noreply.github.com>
Date: Fri, 14 May 2021 01:31:15 +0800
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添加 0700.二叉搜索树中的搜索 Java版本，4种解法
---
 problems/0700.二叉搜索树中的搜索.md | 71 ++++++++++++++++++++
 1 file changed, 71 insertions(+)

diff --git a/problems/0700.二叉搜索树中的搜索.md b/problems/0700.二叉搜索树中的搜索.md
index 5c1cdfdf..401c85d5 100644
--- a/problems/0700.二叉搜索树中的搜索.md
+++ b/problems/0700.二叉搜索树中的搜索.md
@@ -142,7 +142,78 @@ public:
 
 
 Java：
+```Java
+class Solution {
+    // 递归
+    public TreeNode searchBST(TreeNode root, int val) {
+        if (root == null || root.val == val) {
+            return root;
+        }
+        TreeNode left = searchBST(root.left, val);
+        if (left != null) {
+            return left;
+        }
+        return searchBST(root.right, val);
+    }
+}
 
+class Solution {
+    // 递归，利用二叉搜索树特点，优化
+    public TreeNode searchBST(TreeNode root, int val) {
+        if (root == null || root.val == val) {
+            return root;
+        }
+        if (val < root.val) {
+            return searchBST(root.left, val);
+        } else {
+            return searchBST(root.right, val);
+        }
+    }
+}
+
+class Solution {
+    // 迭代
+    public TreeNode searchBST(TreeNode root, int val) {
+        if (root == null || root.val == val) {
+            return root;
+        }
+        Stack<TreeNode> stack = new Stack<>();
+        stack.push(root);
+        while (!stack.isEmpty()) {
+            TreeNode pop = stack.pop();
+            if (pop.val == val) {
+                return pop;
+            }
+            if (pop.right != null) {
+                stack.push(pop.right);
+            }
+            if (pop.left != null) {
+                stack.push(pop.left);
+            }
+        }
+        return null;
+    }
+}
+
+class Solution {
+    // 迭代，利用二叉搜索树特点，优化，可以不需要栈
+    public TreeNode searchBST(TreeNode root, int val) {
+        if (root == null || root.val == val) {
+            return root;
+        }
+        while (root != null) {
+            if (root.val == val) {
+                return root;
+            } else if (val < root.val) {
+                root = root.left;
+            } else {
+                root = root.right;
+            }
+        }
+        return null;
+    }
+}
+```
 
 Python：