Question

1161. 最大层内元素和

English Version

题目描述

给你一个二叉树的根节点 root。设根节点位于二叉树的第 1 层，而根节点的子节点位于第 2 层，依此类推。

请返回层内元素之和 最大 的那几层（可能只有一层）的层号，并返回其中 最小 的那个。

 

示例 1：

输入：root = [1,7,0,7,-8,null,null]
输出：2
解释：
第 1 层各元素之和为 1，
第 2 层各元素之和为 7 + 0 = 7，
第 3 层各元素之和为 7 + -8 = -1，
所以我们返回第 2 层的层号，它的层内元素之和最大。

示例 2：

输入：root = [989,null,10250,98693,-89388,null,null,null,-32127]
输出：2

 

提示：

• 树中的节点数在 [1, 104]范围内
• -105 <= Node.val <= 105

解法

方法一：BFS

BFS 层次遍历，求每一层的节点和，找出节点和最大的层，若有多个层的节点和最大，则返回最小的层。

时间复杂度 $O(n)$，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 maxLevelSum(self, root: Optional[TreeNode]) -> int:
    q = deque([root])
    mx = -inf
    i = 0
    while q:
      i += 1
      s = 0
      for _ in range(len(q)):
        node = q.popleft()
        s += node.val
        if node.left:
          q.append(node.left)
        if node.right:
          q.append(node.right)
      if mx < s:
        mx = s
        ans = i
    return ans

/**
 * 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 maxLevelSum(TreeNode root) {
    Deque<TreeNode> q = new ArrayDeque<>();
    q.offer(root);
    int mx = Integer.MIN_VALUE;
    int i = 0;
    int ans = 0;
    while (!q.isEmpty()) {
      ++i;
      int s = 0;
      for (int n = q.size(); n > 0; --n) {
        TreeNode node = q.pollFirst();
        s += node.val;
        if (node.left != null) {
          q.offer(node.left);
        }
        if (node.right != null) {
          q.offer(node.right);
        }
      }
      if (mx < s) {
        mx = s;
        ans = i;
      }
    }
    return ans;
  }
}

/**
 * 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 maxLevelSum(TreeNode* root) {
    queue<TreeNode*> q{{root}};
    int mx = INT_MIN;
    int ans = 0;
    int i = 0;
    while (!q.empty()) {
      ++i;
      int s = 0;
      for (int n = q.size(); n; --n) {
        root = q.front();
        q.pop();
        s += root->val;
        if (root->left) q.push(root->left);
        if (root->right) q.push(root->right);
      }
      if (mx < s) mx = s, ans = i;
    }
    return ans;
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func maxLevelSum(root *TreeNode) int {
  q := []*TreeNode{root}
  mx := -0x3f3f3f3f
  i := 0
  ans := 0
  for len(q) > 0 {
    i++
    s := 0
    for n := len(q); n > 0; n-- {
      root = q[0]
      q = q[1:]
      s += root.Val
      if root.Left != nil {
        q = append(q, root.Left)
      }
      if root.Right != nil {
        q = append(q, root.Right)
      }
    }
    if mx < s {
      mx = s
      ans = i
    }
  }
  return ans
}

/**
 * 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 maxLevelSum(root: TreeNode | null): number {
  const queue = [root];
  let res = 1;
  let max = -Infinity;
  let h = 1;
  while (queue.length !== 0) {
    const n = queue.length;
    let sum = 0;
    for (let i = 0; i < n; i++) {
      const { val, left, right } = queue.shift();
      sum += val;
      left && queue.push(left);
      right && queue.push(right);
    }
    if (sum > max) {
      max = sum;
      res = h;
    }
    h++;
  }
  return res;
}

方法二：DFS

我们也可以使用 DFS 求解。我们用一个数组 $s$ 来存储每一层的节点和，数组的下标表示层数，数组的值表示节点和。我们使用 DFS 遍历二叉树，将每个节点的值加到对应层数的节点和上。最后，我们返回 $s$ 中的最大值对应的下标即可。

时间复杂度 $O(n)$，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 maxLevelSum(self, root: Optional[TreeNode]) -> int:
    def dfs(node, i):
      if node is None:
        return
      if i == len(s):
        s.append(node.val)
      else:
        s[i] += node.val
      dfs(node.left, i + 1)
      dfs(node.right, i + 1)

    s = []
    dfs(root, 0)
    return s.index(max(s)) + 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<Integer> s = new ArrayList<>();

  public int maxLevelSum(TreeNode root) {
    dfs(root, 0);
    int mx = Integer.MIN_VALUE;
    int ans = 0;
    for (int i = 0; i < s.size(); ++i) {
      if (mx < s.get(i)) {
        mx = s.get(i);
        ans = i + 1;
      }
    }
    return ans;
  }

  private void dfs(TreeNode root, int i) {
    if (root == null) {
      return;
    }
    if (i == s.size()) {
      s.add(root.val);
    } else {
      s.set(i, s.get(i) + root.val);
    }
    dfs(root.left, i + 1);
    dfs(root.right, i + 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:
  int maxLevelSum(TreeNode* root) {
    vector<int> s;
    dfs(root, 0, s);
    int mx = INT_MIN;
    int ans = 0;
    for (int i = 0; i < s.size(); ++i)
      if (mx < s[i]) mx = s[i], ans = i + 1;
    return ans;
  }

  void dfs(TreeNode* root, int i, vector<int>& s) {
    if (!root) return;
    if (s.size() == i)
      s.push_back(root->val);
    else
      s[i] += root->val;
    dfs(root->left, i + 1, s);
    dfs(root->right, i + 1, s);
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func maxLevelSum(root *TreeNode) int {
  s := []int{}
  var dfs func(*TreeNode, int)
  dfs = func(root *TreeNode, i int) {
    if root == nil {
      return
    }
    if len(s) == i {
      s = append(s, root.Val)
    } else {
      s[i] += root.Val
    }
    dfs(root.Left, i+1)
    dfs(root.Right, i+1)
  }
  dfs(root, 0)
  ans, mx := 0, -0x3f3f3f3f
  for i, v := range s {
    if mx < v {
      mx = v
      ans = i + 1
    }
  }
  return ans
}