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feat: add solutions to lc problem: No.0671. Second Minimum Node In a Binary Tree
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‎solution/0600-0699/0671.Second Minimum Node In a Binary Tree/README.md

Lines changed: 53 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -52,15 +52,66 @@
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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class Solution:
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def findSecondMinimumValue(self, root: TreeNode) -> int:
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def traverse(root, val):
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if root is None:
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return
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traverse(root.left, val)
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traverse(root.right, val)
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if root.val > val:
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self.res = root.val if self.res == -1 else min(self.res, root.val)
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self.res = -1
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traverse(root, root.val)
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return self.res
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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/**
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* Definition for a binary tree node.
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* public class TreeNode {
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* int val;
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* TreeNode left;
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* TreeNode right;
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* TreeNode() {}
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* TreeNode(int val) { this.val = val; }
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* TreeNode(int val, TreeNode left, TreeNode right) {
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* this.val = val;
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* this.left = left;
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* this.right = right;
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* }
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* }
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*/
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class Solution {
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private int res;
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public int findSecondMinimumValue(TreeNode root) {
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res = -1;
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traverse(root, root.val);
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return res;
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}
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private void traverse(TreeNode root, int min) {
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if (root == null) {
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return;
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}
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traverse(root.left, min);
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traverse(root.right, min);
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if (root.val > min) {
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res = res == -1 ? root.val : Math.min(res, root.val);
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}
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}
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}
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```
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### **...**

‎solution/0600-0699/0671.Second Minimum Node In a Binary Tree/README_EN.md

Lines changed: 53 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -46,13 +46,64 @@
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### **Python3**
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```python
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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class Solution:
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def findSecondMinimumValue(self, root: TreeNode) -> int:
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def traverse(root, val):
58+
if root is None:
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return
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traverse(root.left, val)
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traverse(root.right, val)
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if root.val > val:
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self.res = root.val if self.res == -1 else min(self.res, root.val)
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self.res = -1
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traverse(root, root.val)
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return self.res
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```
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### **Java**
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```java
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/**
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* Definition for a binary tree node.
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* public class TreeNode {
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* int val;
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* TreeNode left;
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* TreeNode right;
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* TreeNode() {}
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* TreeNode(int val) { this.val = val; }
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* TreeNode(int val, TreeNode left, TreeNode right) {
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* this.val = val;
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* this.left = left;
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* this.right = right;
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* }
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* }
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*/
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class Solution {
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private int res;
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public int findSecondMinimumValue(TreeNode root) {
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res = -1;
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traverse(root, root.val);
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return res;
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}
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private void traverse(TreeNode root, int min) {
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if (root == null) {
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return;
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}
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traverse(root.left, min);
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traverse(root.right, min);
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if (root.val > min) {
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res = res == -1 ? root.val : Math.min(res, root.val);
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}
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}
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}
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```
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### **...**

‎solution/0600-0699/0671.Second Minimum Node In a Binary Tree/Solution.java

Lines changed: 21 additions & 15 deletions
Original file line numberDiff line numberDiff line change
@@ -4,25 +4,31 @@
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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; }
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* TreeNode() {}
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* TreeNode(int val) { this.val = val; }
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* TreeNode(int val, TreeNode left, TreeNode right) {
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* this.val = val;
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* this.left = left;
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* this.right = right;
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* }
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* }
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*/
1016
class Solution {
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private int res;
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public int findSecondMinimumValue(TreeNode root) {
12-
if (root == null || root.left == null) return -1;
13-
int limit = Integer.MAX_VALUE;
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Stack<TreeNode> stack = new Stack<>();
15-
stack.push(root);
16-
while (!stack.isEmpty()) {
17-
TreeNode node = stack.pop();
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if (node.val != root.val) {
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limit = Math.min(limit, node.val - root.val);
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}
21-
if (node.left != null) {
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stack.push(node.left);
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stack.push(node.right);
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}
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res = -1;
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traverse(root, root.val);
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return res;
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}
23+
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private void traverse(TreeNode root, int min) {
25+
if (root == null) {
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return;
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}
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traverse(root.left, min);
29+
traverse(root.right, min);
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if (root.val > min) {
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res = res == -1 ? root.val : Math.min(res, root.val);
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}
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return limit == Integer.MAX_VALUE ? -1 : root.val + limit;
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}
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}

‎solution/0600-0699/0671.Second Minimum Node In a Binary Tree/Solution.py

Lines changed: 19 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,19 @@
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
6+
# self.right = right
7+
class Solution:
8+
def findSecondMinimumValue(self, root: TreeNode) -> int:
9+
def traverse(root, val):
10+
if root is None:
11+
return
12+
traverse(root.left, val)
13+
traverse(root.right, val)
14+
if root.val > val:
15+
self.res = root.val if self.res == -1 else min(self.res, root.val)
16+
17+
self.res = -1
18+
traverse(root, root.val)
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return self.res

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