Question

2583. 二叉树中的第 K 大层和

English Version

题目描述

给你一棵二叉树的根节点 root 和一个正整数 k 。

树中的 层和 是指 同一层 上节点值的总和。

返回树中第 k 大的层和（不一定不同）。如果树少于 k 层，则返回 -1 。

注意，如果两个节点与根节点的距离相同，则认为它们在同一层。

 

示例 1：

输入：root = [5,8,9,2,1,3,7,4,6], k = 2
输出：13
解释：树中每一层的层和分别是：
- Level 1: 5
- Level 2: 8 + 9 = 17
- Level 3: 2 + 1 + 3 + 7 = 13
- Level 4: 4 + 6 = 10
第 2 大的层和等于 13 。

示例 2：

输入：root = [1,2,null,3], k = 1
输出：3
解释：最大的层和是 3 。

 

提示：

• 树中的节点数为 n
• 2 <= n <= 105
• 1 <= Node.val <= 106
• 1 <= k <= n

解法

方法一：BFS + 排序

我们可以使用 BFS 遍历二叉树，同时记录每一层的节点和，然后对节点和数组进行排序，最后返回第 $k$ 大的节点和即可。注意，如果二叉树的层数小于 $k$，则返回 $-1$。

时间复杂度 $O(n \times \log n)$，空间复杂度 $O(n)$。其中 $n$ 为二叉树的节点数。

# 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 kthLargestLevelSum(self, root: Optional[TreeNode], k: int) -> int:
    arr = []
    q = deque([root])
    while q:
      t = 0
      for _ in range(len(q)):
        root = q.popleft()
        t += root.val
        if root.left:
          q.append(root.left)
        if root.right:
          q.append(root.right)
      arr.append(t)
    return -1 if len(arr) < k else nlargest(k, arr)[-1]

/**
 * 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 long kthLargestLevelSum(TreeNode root, int k) {
    List<Long> arr = new ArrayList<>();
    Deque<TreeNode> q = new ArrayDeque<>();
    q.offer(root);
    while (!q.isEmpty()) {
      long t = 0;
      for (int n = q.size(); n > 0; --n) {
        root = q.pollFirst();
        t += root.val;
        if (root.left != null) {
          q.offer(root.left);
        }
        if (root.right != null) {
          q.offer(root.right);
        }
      }
      arr.add(t);
    }
    if (arr.size() < k) {
      return -1;
    }
    Collections.sort(arr, Collections.reverseOrder());
    return arr.get(k - 1);
  }
}

/**
 * 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:
  long long kthLargestLevelSum(TreeNode* root, int k) {
    vector<long long> arr;
    queue<TreeNode*> q{{root}};
    while (!q.empty()) {
      long long t = 0;
      for (int n = q.size(); n; --n) {
        root = q.front();
        q.pop();
        t += root->val;
        if (root->left) {
          q.push(root->left);
        }
        if (root->right) {
          q.push(root->right);
        }
      }
      arr.push_back(t);
    }
    if (arr.size() < k) {
      return -1;
    }
    sort(arr.rbegin(), arr.rend());
    return arr[k - 1];
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func kthLargestLevelSum(root *TreeNode, k int) int64 {
  arr := []int{}
  q := []*TreeNode{root}
  for len(q) > 0 {
    t := 0
    for n := len(q); n > 0; n-- {
      root = q[0]
      q = q[1:]
      t += root.Val
      if root.Left != nil {
        q = append(q, root.Left)
      }
      if root.Right != nil {
        q = append(q, root.Right)
      }
    }
    arr = append(arr, t)
  }
  if n := len(arr); n >= k {
    sort.Ints(arr)
    return int64(arr[n-k])
  }
  return -1
}

/**
 * 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 kthLargestLevelSum(root: TreeNode | null, k: number): number {
  const arr: number[] = [];
  const q = [root];
  while (q.length) {
    let t = 0;
    for (let n = q.length; n > 0; --n) {
      root = q.shift();
      t += root.val;
      if (root.left) {
        q.push(root.left);
      }
      if (root.right) {
        q.push(root.right);
      }
    }
    arr.push(t);
  }
  if (arr.length < k) {
    return -1;
  }
  arr.sort((a, b) => b - a);
  return arr[k - 1];
}

方法二：DFS + 排序

我们也可以使用 DFS 遍历二叉树，同时记录每一层的节点和，然后对节点和数组进行排序，最后返回第 $k$ 大的节点和即可。注意，如果二叉树的层数小于 $k$，则返回 $-1$。

时间复杂度 $O(n \times \log n)$，空间复杂度 $O(n)$。其中 $n$ 为二叉树的节点数。

# 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 kthLargestLevelSum(self, root: Optional[TreeNode], k: int) -> int:
    def dfs(root, d):
      if root is None:
        return
      if len(arr) <= d:
        arr.append(0)
      arr[d] += root.val
      dfs(root.left, d + 1)
      dfs(root.right, d + 1)

    arr = []
    dfs(root, 0)
    return -1 if len(arr) < k else nlargest(k, arr)[-1]

/**
 * 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 {
  private List<Long> arr = new ArrayList<>();

  public long kthLargestLevelSum(TreeNode root, int k) {
    dfs(root, 0);
    if (arr.size() < k) {
      return -1;
    }
    Collections.sort(arr, Collections.reverseOrder());
    return arr.get(k - 1);
  }

  private void dfs(TreeNode root, int d) {
    if (root == null) {
      return;
    }
    if (arr.size() <= d) {
      arr.add(0L);
    }
    arr.set(d, arr.get(d) + root.val);
    dfs(root.left, d + 1);
    dfs(root.right, d + 1);
  }
}

/**
 * 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:
  long long kthLargestLevelSum(TreeNode* root, int k) {
    vector<long long> arr;
    function<void(TreeNode*, int)> dfs = [&](TreeNode* root, int d) {
      if (!root) {
        return;
      }
      if (arr.size() <= d) {
        arr.push_back(0);
      }
      arr[d] += root->val;
      dfs(root->left, d + 1);
      dfs(root->right, d + 1);
    };
    dfs(root, 0);
    if (arr.size() < k) {
      return -1;
    }
    sort(arr.rbegin(), arr.rend());
    return arr[k - 1];
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func kthLargestLevelSum(root *TreeNode, k int) int64 {
  arr := []int{}
  var dfs func(*TreeNode, int)
  dfs = func(root *TreeNode, d int) {
    if root == nil {
      return
    }
    if len(arr) <= d {
      arr = append(arr, 0)
    }
    arr[d] += root.Val
    dfs(root.Left, d+1)
    dfs(root.Right, d+1)
  }

  dfs(root, 0)
  if n := len(arr); n >= k {
    sort.Ints(arr)
    return int64(arr[n-k])
  }
  return -1
}

/**
 * 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 kthLargestLevelSum(root: TreeNode | null, k: number): number {
  const dfs = (root: TreeNode, d: number) => {
    if (!root) {
      return;
    }
    if (arr.length <= d) {
      arr.push(0);
    }
    arr[d] += root.val;
    dfs(root.left, d + 1);
    dfs(root.right, d + 1);
  };
  const arr: number[] = [];
  dfs(root, 0);
  if (arr.length < k) {
    return -1;
  }
  arr.sort((a, b) => b - a);
  return arr[k - 1];
}