feat: add solutions to lc problem: No.3157 (doocs#2888)

No.3157.Find the Level of Tree with Minimum Sum

Author: yanglbme

solution/3100-3199/3157.Find the Level of Tree with Minimum Sum/README.md

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+---
+comments: true
+difficulty: 中等
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3157.Find%20the%20Level%20of%20Tree%20with%20Minimum%20Sum/README.md
+---
+
+<!-- problem:start -->
+
+# [3157. Find the Level of Tree with Minimum Sum 🔒](https://leetcode.cn/problems/find-the-level-of-tree-with-minimum-sum)
+
+[English Version](/solution/3100-3199/3157.Find%20the%20Level%20of%20Tree%20with%20Minimum%20Sum/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>Given the root of a binary tree <code>root</code> where each node has a value, return the level of the tree that has the <strong>minimum</strong> sum of values among all the levels (in case of a tie, return the <strong>lowest</strong> level).</p>
+
+<p><strong>Note</strong> that the root of the tree is at level 1 and the level of any other node is its distance from the root + 1.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">root = [50,6,2,30,80,7]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">2</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3100-3199/3157.Find%20the%20Level%20of%20Tree%20with%20Minimum%20Sum/images/image_2024-05-17_16-15-46.png" style="padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; width: 265px; height: 129px;" /></p>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">root = [36,17,10,null,null,24]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">3</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3100-3199/3157.Find%20the%20Level%20of%20Tree%20with%20Minimum%20Sum/images/image_2024-05-17_16-14-18.png" style="padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; width: 170px; height: 135px;" /></p>
+</div>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">root = [5,null,5,null,5]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">1</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3100-3199/3157.Find%20the%20Level%20of%20Tree%20with%20Minimum%20Sum/images/image_2024-05-19_19-07-20.png" style="padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; width: 170px; height: 135px;" /></p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li>The number of nodes in the tree is in the range <code>[1, 10<sup>5</sup>]</code>.</li>
+	<li><code>1 &lt;= Node.val &lt;= 10<sup>9</sup></code></li>
+</ul>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一：BFS
+
+我们可以使用 BFS，逐层遍历二叉树，记录每一层的节点值之和，找到具有最小节点值之和的层，返回该层的层数。
+
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为二叉树的节点个数。
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+class Solution:
+    def minimumLevel(self, root: Optional[TreeNode]) -> int:
+        q = deque([root])
+        ans = 0
+        level, s = 1, inf
+        while q:
+            t = 0
+            for _ in range(len(q)):
+                node = q.popleft()
+                t += node.val
+                if node.left:
+                    q.append(node.left)
+                if node.right:
+                    q.append(node.right)
+            if s > t:
+                s = t
+                ans = level
+            level += 1
+        return ans
+```
+
+#### Java
+
+```java
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode() {}
+ *     TreeNode(int val) { this.val = val; }
+ *     TreeNode(int val, TreeNode left, TreeNode right) {
+ *         this.val = val;
+ *         this.left = left;
+ *         this.right = right;
+ *     }
+ * }
+ */
+class Solution {
+    public int minimumLevel(TreeNode root) {
+        Deque<TreeNode> q = new ArrayDeque<>();
+        q.offer(root);
+        int ans = 0;
+        long s = Long.MAX_VALUE;
+        for (int level = 1; !q.isEmpty(); ++level) {
+            long t = 0;
+            for (int m = q.size(); m > 0; --m) {
+                TreeNode node = q.poll();
+                t += node.val;
+                if (node.left != null) {
+                    q.offer(node.left);
+                }
+                if (node.right != null) {
+                    q.offer(node.right);
+                }
+            }
+            if (s > t) {
+                s = t;
+                ans = level;
+            }
+        }
+        return ans;
+    }
+}
+```
+
+#### C++
+
+```cpp
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
+ * };
+ */
+class Solution {
+public:
+    int minimumLevel(TreeNode* root) {
+        queue<TreeNode*> q{{root}};
+        int ans = 0;
+        long long s = 1LL << 60;
+        for (int level = 1; q.size(); ++level) {
+            long long t = 0;
+            for (int m = q.size(); m; --m) {
+                TreeNode* node = q.front();
+                q.pop();
+                t += node->val;
+                if (node->left) {
+                    q.push(node->left);
+                }
+                if (node->right) {
+                    q.push(node->right);
+                }
+            }
+            if (s > t) {
+                s = t;
+                ans = level;
+            }
+        }
+        return ans;
+    }
+};
+```
+
+#### Go
+
+```go
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+func minimumLevel(root *TreeNode) (ans int) {
+	q := []*TreeNode{root}
+	s := math.MaxInt64
+	for level := 1; len(q) > 0; level++ {
+		t := 0
+		for m := len(q); m > 0; m-- {
+			node := q[0]
+			q = q[1:]
+			t += node.Val
+			if node.Left != nil {
+				q = append(q, node.Left)
+			}
+			if node.Right != nil {
+				q = append(q, node.Right)
+			}
+		}
+		if s > t {
+			s = t
+			ans = level
+		}
+	}
+	return
+}
+```
+
+#### TypeScript
+
+```ts
+/**
+ * Definition for a binary tree node.
+ * class TreeNode {
+ *     val: number
+ *     left: TreeNode | null
+ *     right: TreeNode | null
+ *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
+ *         this.val = (val===undefined ? 0 : val)
+ *         this.left = (left===undefined ? null : left)
+ *         this.right = (right===undefined ? null : right)
+ *     }
+ * }
+ */
+
+function minimumLevel(root: TreeNode | null): number {
+    const q: TreeNode[] = [root];
+    let s = Infinity;
+    let ans = 0;
+    for (let level = 1; q.length; ++level) {
+        const qq: TreeNode[] = [];
+        let t = 0;
+        for (const { val, left, right } of q) {
+            t += val;
+            left && qq.push(left);
+            right && qq.push(right);
+        }
+        if (s > t) {
+            s = t;
+            ans = level;
+        }
+        q.splice(0, q.length, ...qq);
+    }
+    return ans;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->