Question

1302. 层数最深叶子节点的和

English Version

题目描述

给你一棵二叉树的根节点 root ，请你返回 层数最深的叶子节点的和 。

 

示例 1：

输入：root = [1,2,3,4,5,null,6,7,null,null,null,null,8]
输出：15

示例 2：

输入：root = [6,7,8,2,7,1,3,9,null,1,4,null,null,null,5]
输出：19

 

提示：

• 树中节点数目在范围 [1, 104] 之间。
• 1 <= Node.val <= 100

解法

方法一：BFS

可以忽略一些细节，每次都统计当前遍历层级的数值和，当 BFS 结束时，最后一次数值和便是结果。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是树中节点的数目。

# 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 deepestLeavesSum(self, root: Optional[TreeNode]) -> int:
    q = deque([root])
    while q:
      ans = 0
      for _ in range(len(q)):
        root = q.popleft()
        ans += root.val
        if root.left:
          q.append(root.left)
        if root.right:
          q.append(root.right)
    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 deepestLeavesSum(TreeNode root) {
    Deque<TreeNode> q = new ArrayDeque<>();
    q.offer(root);
    int ans = 0;
    while (!q.isEmpty()) {
      ans = 0;
      for (int n = q.size(); n > 0; --n) {
        root = q.pollFirst();
        ans += root.val;
        if (root.left != null) {
          q.offer(root.left);
        }
        if (root.right != null) {
          q.offer(root.right);
        }
      }
    }
    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 deepestLeavesSum(TreeNode* root) {
    int ans = 0;
    queue<TreeNode*> q{{root}};
    while (!q.empty()) {
      ans = 0;
      for (int n = q.size(); n; --n) {
        root = q.front();
        q.pop();
        ans += root->val;
        if (root->left) q.push(root->left);
        if (root->right) q.push(root->right);
      }
    }
    return ans;
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func deepestLeavesSum(root *TreeNode) int {
  q := []*TreeNode{root}
  ans := 0
  for len(q) > 0 {
    ans = 0
    for n := len(q); n > 0; n-- {
      root = q[0]
      q = q[1:]
      ans += root.Val
      if root.Left != nil {
        q = append(q, root.Left)
      }
      if root.Right != nil {
        q = append(q, root.Right)
      }
    }
  }
  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 deepestLeavesSum(root: TreeNode | null): number {
  const queue = [root];
  let res = 0;
  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);
    }
    res = sum;
  }
  return res;
}

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//   TreeNode {
//     val,
//     left: None,
//     right: None
//   }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
  fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, depth: i32, max_depth: &mut i32, res: &mut i32) {
    if let Some(node) = root {
      let node = node.borrow();
      if node.left.is_none() && node.right.is_none() {
        if depth == *max_depth {
          *res += node.val;
        } else if depth > *max_depth {
          *max_depth = depth;
          *res = node.val;
        }
        return;
      }
      Self::dfs(&node.left, depth + 1, max_depth, res);
      Self::dfs(&node.right, depth + 1, max_depth, res);
    }
  }

  pub fn deepest_leaves_sum(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
    let mut res = 0;
    let mut max_depth = 0;
    Self::dfs(&root, 0, &mut max_depth, &mut res);
    res
  }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *   int val;
 *   struct TreeNode *left;
 *   struct TreeNode *right;
 * };
 */

void dfs(struct TreeNode* root, int depth, int* maxDepth, int* res) {
  if (!root->left && !root->right) {
    if (depth == *maxDepth) {
      *res += root->val;
    } else if (depth > *maxDepth) {
      *maxDepth = depth;
      *res = root->val;
    }
    return;
  }
  if (root->left) {
    dfs(root->left, depth + 1, maxDepth, res);
  }
  if (root->right) {
    dfs(root->right, depth + 1, maxDepth, res);
  }
}

int deepestLeavesSum(struct TreeNode* root) {
  int res = 0;
  int maxDepth = 0;
  dfs(root, 0, &maxDepth, &res);
  return res;
}

方法二：DFS

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是树中节点的数目。

# 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 deepestLeavesSum(self, root: Optional[TreeNode]) -> int:
    def dfs(root, i):
      nonlocal ans, mx
      if root is None:
        return
      if i == mx:
        ans += root.val
      elif i > mx:
        ans = root.val
        mx = i
      dfs(root.left, i + 1)
      dfs(root.right, i + 1)

    ans = mx = 0
    dfs(root, 1)
    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 {
  int mx;
  int ans;

  public int deepestLeavesSum(TreeNode root) {
    dfs(root, 1);
    return ans;
  }

  private void dfs(TreeNode root, int i) {
    if (root == null) {
      return;
    }
    if (i > mx) {
      mx = i;
      ans = root.val;
    } else if (i == mx) {
      ans += 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 mx = 0;
  int ans = 0;

  int deepestLeavesSum(TreeNode* root) {
    dfs(root, 1);
    return ans;
  }

  void dfs(TreeNode* root, int i) {
    if (!root) return;
    if (i == mx) {
      ans += root->val;
    } else if (i > mx) {
      mx = i;
      ans = root->val;
    }
    dfs(root->left, i + 1);
    dfs(root->right, i + 1);
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func deepestLeavesSum(root *TreeNode) int {
  ans, mx := 0, 0
  var dfs func(*TreeNode, int)
  dfs = func(root *TreeNode, i int) {
    if root == nil {
      return
    }
    if i == mx {
      ans += root.Val
    } else if i > mx {
      mx = i
      ans = root.Val
    }
    dfs(root.Left, i+1)
    dfs(root.Right, i+1)
  }
  dfs(root, 1)
  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 deepestLeavesSum(root: TreeNode | null): number {
  let res = 0;
  let maxDepath = 0;
  const dfs = ({ val, left, right }: TreeNode, depth: number) => {
    if (left == null && right == null) {
      if (depth === maxDepath) {
        res += val;
      } else if (depth > maxDepath) {
        maxDepath = depth;
        res = val;
      }
      return;
    }
    left && dfs(left, depth + 1);
    right && dfs(right, depth + 1);
  };
  dfs(root, 0);
  return res;
}