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3157. 找到具有最小和的树的层数 🔒

题目描述

给定一棵二叉树的根 root，其中每个节点有一个值，返回树中 层和最小 的层数（如果相等，返回 最低 的层数）。

注意 树的根节点在第一层，其它任何节点的层数是它到根节点的距离+1。

 

示例 1：

输入：root = [50,6,2,30,80,7]

输出：2

解释：

示例 2：

输入：root = [36,17,10,null,null,24]

输出：3

解释：

示例 3：

输入：root = [5,null,5,null,5]

输出：1

解释：

 

提示：

  • 树中节点数量的范围是 [1, 105]。
  • 1 <= Node.val <= 109

解法

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。

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# 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

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/**
 * 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;
    }
}

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/**
 * 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;
    }
};

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/**
 * 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
}

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/**
 * 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;
}

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