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Commit 15d8d08

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feat: add solutions to lc/lcof2 problems (doocs#1404)
* lc No.0285.Inorder Successor in BST * lcof2 No.053.二叉搜索树中的中序后继
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‎lcof2/剑指 Offer II 053. 二叉搜索树中的中序后继/README.md‎

Lines changed: 85 additions & 41 deletions
Original file line numberDiff line numberDiff line change
@@ -48,7 +48,18 @@
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<!-- 这里可写通用的实现逻辑 -->
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51-
利用二叉搜索树的特性,`p` 的中序后继一定是所有大于 `p` 的节点中最小的那个
51+
**方法一:二分搜索**
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二叉搜索树的中序遍历是一个升序序列,因此可以使用二分搜索的方法。
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二叉搜索树节点 $p$ 的中序后继节点满足:
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1. 中序后继的节点值大于 $p$ 的节点值
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2. 中序后继是所有大于 $p$ 的节点中值最小的节点
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因此,对于当前节点 $root,ドル如果 $root.val \gt p.val,ドル则 $root$ 可能是 $p$ 的中序后继节点,将 $root$ 记为 $ans,ドル然后搜索左子树,即 $root = root.left$;如果 $root.val \leq p.val,ドル则 $root$ 不能是 $p$ 的中序后继节点,搜索右子树,即 $root = root.right$。
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时间复杂度 $O(h),ドル其中 $h$ 为二叉搜索树的高度。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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@@ -66,14 +77,14 @@
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class Solution:
69-
def inorderSuccessor(self, root: 'TreeNode', p: 'TreeNode') -> 'TreeNode':
70-
cur, ans = root, None
71-
while cur:
72-
if cur.val <= p.val:
73-
cur = cur.right
80+
def inorderSuccessor(self, root: "TreeNode", p: "TreeNode") -> "TreeNode":
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ans = None
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while root:
83+
if root.val > p.val:
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ans = root
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root = root.left
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else:
75-
ans = cur
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cur = cur.left
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root = root.right
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return ans
7889
```
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@@ -93,20 +104,49 @@ class Solution:
93104
*/
94105
class Solution {
95106
public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
96-
TreeNode cur = root, ans = null;
97-
while (cur != null) {
98-
if (cur.val <= p.val) {
99-
cur = cur.right;
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TreeNode ans = null;
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while (root != null) {
109+
if (root.val > p.val) {
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ans = root;
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root = root.left;
100112
} else {
101-
ans = cur;
102-
cur = cur.left;
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root = root.right;
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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* };
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*/
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class Solution {
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public:
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TreeNode* inorderSuccessor(TreeNode* root, TreeNode* p) {
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TreeNode* ans = nullptr;
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while (root) {
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if (root->val > p->val) {
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ans = root;
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root = root->left;
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} else {
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root = root->right;
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}
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
@@ -119,46 +159,50 @@ class Solution {
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* }
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*/
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func inorderSuccessor(root *TreeNode, p *TreeNode) (ans *TreeNode) {
122-
cur:= root
123-
for cur != nil {
124-
if cur.Val <= p.Val {
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cur = cur.Right
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for root != nil {
163+
if root.Val > p.Val {
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ans = root
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root = root.Left
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} else {
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ans = cur
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cur = cur.Left
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root = root.Right
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}
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}
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return
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}
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```
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135-
### **C++**
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### **TypeScript**
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137-
```cpp
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```ts
138177
/**
139178
* Definition for a binary tree node.
140-
* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
145-
* };
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* class TreeNode {
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* val: number
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* left: TreeNode | null
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* right: TreeNode | null
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* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
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* this.val = (val===undefined ? 0 : val)
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* this.left = (left===undefined ? null : left)
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* this.right = (right===undefined ? null : right)
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* }
188+
* }
146189
*/
147-
class Solution {
148-
public:
149-
TreeNode* inorderSuccessor(TreeNode* root, TreeNode* p) {
150-
TreeNode *cur = root, *ans = nullptr;
151-
while (cur != nullptr) {
152-
if (cur->val <= p->val) {
153-
cur = cur->right;
154-
} else {
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ans = cur;
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cur = cur->left;
157-
}
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191+
function inorderSuccessor(
192+
root: TreeNode | null,
193+
p: TreeNode | null,
194+
): TreeNode | null {
195+
let ans: TreeNode | null = null;
196+
while (root) {
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if (root.val > p.val) {
198+
ans = root;
199+
root = root.left;
200+
} else {
201+
root = root.right;
158202
}
159-
return ans;
160203
}
161-
};
204+
return ans;
205+
}
162206
```
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164208
### **...**
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Original file line numberDiff line numberDiff line change
@@ -1,24 +1,24 @@
1-
/**
2-
* Definition for a binary tree node.
3-
* struct TreeNode {
4-
* int val;
5-
* TreeNode *left;
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* TreeNode *right;
7-
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
8-
* };
9-
*/
10-
class Solution {
11-
public:
12-
TreeNode* inorderSuccessor(TreeNode* root, TreeNode* p) {
13-
TreeNode *cur = root, *ans = nullptr;
14-
while (cur != nullptr) {
15-
if (cur->val <= p->val) {
16-
cur = cur->right;
17-
} else {
18-
ans = cur;
19-
cur = cur->left;
20-
}
21-
}
22-
return ans;
23-
}
24-
};
1+
/**
2+
* Definition for a binary tree node.
3+
* struct TreeNode {
4+
* int val;
5+
* TreeNode *left;
6+
* TreeNode *right;
7+
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
8+
* };
9+
*/
10+
class Solution {
11+
public:
12+
TreeNode* inorderSuccessor(TreeNode* root, TreeNode* p) {
13+
TreeNode* ans = nullptr;
14+
while (root) {
15+
if (root->val > p->val) {
16+
ans = root;
17+
root = root->left;
18+
} else {
19+
root = root->right;
20+
}
21+
}
22+
return ans;
23+
}
24+
};

‎lcof2/剑指 Offer II 053. 二叉搜索树中的中序后继/Solution.go‎

Lines changed: 5 additions & 6 deletions
Original file line numberDiff line numberDiff line change
@@ -7,13 +7,12 @@
77
* }
88
*/
99
func inorderSuccessor(root *TreeNode, p *TreeNode) (ans *TreeNode) {
10-
cur:=root
11-
forcur!=nil {
12-
ifcur.Val<=p.Val {
13-
cur = cur.Right
10+
forroot!=nil {
11+
ifroot.Val>p.Val {
12+
ans=root
13+
root = root.Left
1414
} else {
15-
ans = cur
16-
cur = cur.Left
15+
root = root.Right
1716
}
1817
}
1918
return
Lines changed: 23 additions & 23 deletions
Original file line numberDiff line numberDiff line change
@@ -1,23 +1,23 @@
1-
/**
2-
* Definition for a binary tree node.
3-
* public class TreeNode {
4-
* int val;
5-
* TreeNode left;
6-
* TreeNode right;
7-
* TreeNode(int x) { val = x; }
8-
* }
9-
*/
10-
class Solution {
11-
public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
12-
TreeNode cur = root, ans = null;
13-
while (cur != null) {
14-
if (cur.val <= p.val) {
15-
cur = cur.right;
16-
} else {
17-
ans = cur;
18-
cur = cur.left;
19-
}
20-
}
21-
return ans;
22-
}
23-
}
1+
/**
2+
* Definition for a binary tree node.
3+
* public class TreeNode {
4+
* int val;
5+
* TreeNode left;
6+
* TreeNode right;
7+
* TreeNode(int x) { val = x; }
8+
* }
9+
*/
10+
class Solution {
11+
public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
12+
TreeNode ans = null;
13+
while (root != null) {
14+
if (root.val > p.val) {
15+
ans = root;
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root = root.left;
17+
} else {
18+
root = root.right;
19+
}
20+
}
21+
return ans;
22+
}
23+
}
Lines changed: 18 additions & 18 deletions
Original file line numberDiff line numberDiff line change
@@ -1,18 +1,18 @@
1-
# Definition for a binary tree node.
2-
# class TreeNode:
3-
# def __init__(self, x):
4-
# self.val = x
5-
# self.left = None
6-
# self.right = None
7-
8-
9-
class Solution:
10-
def inorderSuccessor(self, root: 'TreeNode', p: 'TreeNode') -> 'TreeNode':
11-
cur, ans =root, None
12-
while cur:
13-
if cur.val <= p.val:
14-
cur = cur.right
15-
else:
16-
ans=cur
17-
cur = cur.left
18-
return ans
1+
# Definition for a binary tree node.
2+
# class TreeNode:
3+
# def __init__(self, x):
4+
# self.val = x
5+
# self.left = None
6+
# self.right = None
7+
8+
9+
class Solution:
10+
def inorderSuccessor(self, root: "TreeNode", p: "TreeNode") -> "TreeNode":
11+
ans = None
12+
while root:
13+
if root.val > p.val:
14+
ans = root
15+
root=root.left
16+
else:
17+
root = root.right
18+
return ans
Lines changed: 29 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,29 @@
1+
/**
2+
* Definition for a binary tree node.
3+
* class TreeNode {
4+
* val: number
5+
* left: TreeNode | null
6+
* right: TreeNode | null
7+
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
8+
* this.val = (val===undefined ? 0 : val)
9+
* this.left = (left===undefined ? null : left)
10+
* this.right = (right===undefined ? null : right)
11+
* }
12+
* }
13+
*/
14+
15+
function inorderSuccessor(
16+
root: TreeNode | null,
17+
p: TreeNode | null,
18+
): TreeNode | null {
19+
let ans: TreeNode | null = null;
20+
while (root) {
21+
if (root.val > p.val) {
22+
ans = root;
23+
root = root.left;
24+
} else {
25+
root = root.right;
26+
}
27+
}
28+
return ans;
29+
}

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