in order successor in bst

src/main/java/lintcode/medium/LintCode_0448_InorderSuccessorInBST.java

-package lintcode.medium;
-
-import java.util.ArrayList;
-import java.util.List;
-import java.util.Stack;
-
-// leetcode 285
-// https://www.lintcode.com/problem/inorder-successor-in-bst/description
-public class LintCode_0448_InorderSuccessorInBST {
-	public class TreeNode {
-		int val;
-		TreeNode left;
-		TreeNode right;
-
-		TreeNode(int x) {
-			val = x;
-		}
-	}
-
-	public static TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
-		if (p == null) {
-			return null;
-		}
-		if (p.right != null) {
-			return rightLeftMost(p.right);
-		}
-		TreeNode sucessor = null;
-		while (root != null) {
-			if (root.val > p.val) {
-				sucessor = root;
-				root = root.left;
-			} else if (root.val < p.val) {
-				root = root.right;
-			} else {
-				break;
-			}
-		}
-		return sucessor;
-	}
-
-	private static TreeNode rightLeftMost(TreeNode p) {
-		while (p.left != null) {
-			p = p.left;
-		}
-		return p;
-	}
-
-	public static TreeNode inorderSuccessor2(TreeNode root, TreeNode p) {
-		List<TreeNode> ans = new ArrayList<>();
-		if (root == null) {
-			return null;
-		}
-		in2(root, ans);
-		boolean find = false;
-		for (TreeNode c : ans) {
-			if (c == p) {
-				find = true;
-			} else if (find) {
-				return c;
-			}
-		}
-		return null;
-	}
-
-	private static void in2(TreeNode root, List<TreeNode> ans) {
-		if (root == null) {
-			return;
-		}
-		in2(root.left, ans);
-		ans.add(root);
-		in2(root.right, ans);
-	}
-
-	public TreeNode inorderSuccessor3(TreeNode root, TreeNode p) {
-		if (root == null) {
-			return null;
-		}
-		boolean flag = false;
-		Stack<TreeNode> stack = new Stack<>();
-		TreeNode cur = root;
-		while (!stack.isEmpty() || cur != null) {
-			if (cur != null) {
-				stack.push(cur);
-				cur = cur.left;
-			} else {
-				cur = stack.pop();
-				if (cur == p) {
-					flag = true;
-				} else if (flag) {
-					return cur;
-				}
-				cur = cur.right;
-			}
-		}
-		return null;
-	}
-}
+package lintcode.medium;
+
+import java.util.ArrayList;
+import java.util.List;
+import java.util.Stack;
+
+// 笔记：https://www.cnblogs.com/greyzeng/p/16863345.html
+// 二叉搜索树中查找后继节点
+// leetcode 285
+// https://www.lintcode.com/problem/inorder-successor-in-bst/description
+public class LintCode_0448_InorderSuccessorInBST {
+    public class TreeNode {
+        int val;
+        TreeNode left;
+        TreeNode right;
+
+        TreeNode(int x) {
+            val = x;
+        }
+    }
+
+    public static TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
+        if (p == null) {
+            return null;
+        }
+        if (p.right != null) {
+            return rightLeftMost(p.right);
+        }
+        TreeNode successor = null;
+        while (root != null) {
+            if (root.val > p.val) {
+                successor = root;
+                root = root.left;
+            } else if (root.val < p.val) {
+                root = root.right;
+            } else {
+                break;
+            }
+        }
+        return successor;
+    }
+
+    private static TreeNode rightLeftMost(TreeNode p) {
+        while (p.left != null) {
+            p = p.left;
+        }
+        return p;
+    }
+
+    public static TreeNode inorderSuccessor2(TreeNode root, TreeNode p) {
+        List<TreeNode> ans = new ArrayList<>();
+        if (root == null) {
+            return null;
+        }
+        in2(root, ans);
+        boolean find = false;
+        for (TreeNode c : ans) {
+            if (c == p) {
+                find = true;
+            } else if (find) {
+                return c;
+            }
+        }
+        return null;
+    }
+
+    private static void in2(TreeNode root, List<TreeNode> ans) {
+        if (root == null) {
+            return;
+        }
+        in2(root.left, ans);
+        ans.add(root);
+        in2(root.right, ans);
+    }
+
+    public TreeNode inorderSuccessor4(TreeNode head, TreeNode p) {
+        if (head == null) {
+            return null;
+        }
+        TreeNode ans = null;
+        TreeNode cur = head;
+        TreeNode mostRight;
+        boolean find = false;
+        while (cur != null) {
+            mostRight = cur.left;
+            if (mostRight != null) {
+                while (mostRight.right != null && mostRight.right != cur) {
+                    mostRight = mostRight.right;
+                }
+                if (mostRight.right == null) {
+                    mostRight.right = cur;
+                    cur = cur.left;
+                    continue;
+                } else {
+                    mostRight.right = null;
+                }
+            }
+            if (find) {
+                ans = cur;
+                find = false;
+            }
+            if (cur == p) {
+                find = true;
+            }
+            cur = cur.right;
+        }
+        return ans;
+    }
+
+    public TreeNode inorderSuccessor3(TreeNode root, TreeNode p) {
+        if (root == null) {
+            return null;
+        }
+        boolean flag = false;
+        Stack<TreeNode> stack = new Stack<>();
+        TreeNode cur = root;
+        while (!stack.isEmpty() || cur != null) {
+            if (cur != null) {
+                stack.push(cur);
+                cur = cur.left;
+            } else {
+                cur = stack.pop();
+                if (cur == p) {
+                    flag = true;
+                } else if (flag) {
+                    return cur;
+                }
+                cur = cur.right;
+            }
+        }
+        return null;
+    }
+}

src/main/java/snippet/Code_0085_Coffee.java → src/main/java/snippet/Code_MakeCoffee.java

@@ -17,7 +17,7 @@ import java.util.PriorityQueue;
 // 思路：
 // 第一步：先想办法得到一个数组，获取每个人喝咖啡的最少时间数组（小根堆  能用时间+多久制作的时间）
 // 第二步：process(i,free) 从i号咖啡杯开始，一直到最后，最短时间是？其中free是咖啡杯可用的时间
-public class Code_0085_Coffee {
+public class Code_MakeCoffee {
 
 
     public static int minTime0(int[] arr, int n, int wash, int air) {