From fb03de4c4ed82d239102718565c4734e01a211ac Mon Sep 17 00:00:00 2001
From: carlvine500 <carlvine500@163.com>
Date: Tue, 2 Apr 2024 17:46:11 +0800
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 =?UTF-8?q?=E7=9A=84=E6=9C=80=E8=BF=91=E5=85=AC=E5=85=B1=E7=A5=96=E5=85=88?=
 =?UTF-8?q?.md?=
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0236 java简洁迭代解法
---
 .../0236.二叉树的最近公共祖先.md    | 43 ++++++++++++++++++-
 1 file changed, 42 insertions(+), 1 deletion(-)

diff --git a/problems/0236.二叉树的最近公共祖先.md b/problems/0236.二叉树的最近公共祖先.md
index 049f70c7..7e0a12fa 100644
--- a/problems/0236.二叉树的最近公共祖先.md
+++ b/problems/0236.二叉树的最近公共祖先.md
@@ -247,7 +247,7 @@ public:
 
 
 ### Java 
-
+递归
 ```Java
 class Solution {
     public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
@@ -271,6 +271,47 @@ class Solution {
     }
 }
 
+```
+迭代
+```Java
+class Solution {
+    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
+        int max = Integer.MAX_VALUE;
+        Stack<TreeNode> st = new Stack<>();
+        TreeNode cur = root, pre = null;
+        while (cur != null || !st.isEmpty()) {
+            while (cur != null) {
+                st.push(cur);
+                cur = cur.left;
+            }
+            cur = st.pop();
+            if (cur.right == null || cur.right == pre) {
+                // p/q是 中/左 或者 中/右 , 返回中
+                if (cur == p || cur == q) {
+                    if ((cur.left != null && cur.left.val == max) || (cur.right != null && cur.right.val == max)) {
+                        return cur;
+                    }
+                    cur.val = max;
+                }
+                // p/q是 左/右 , 返回中
+                if (cur.left != null && cur.left.val == max && cur.right != null && cur.right.val == max) {
+                    return cur;
+                }
+                // MAX_VALUE 往上传递
+                if ((cur.left != null && cur.left.val == max) || (cur.right != null && cur.right.val == max)) {
+                    cur.val = max;
+                }
+                pre = cur;
+                cur = null;
+            } else {
+                st.push(cur);
+                cur = cur.right;
+            }
+        }
+        return null;
+    }
+}
+
 ```
 
 ### Python