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1302. Deepest Leaves Sum

Description

Given the root of a binary tree, return the sum of values of its deepest leaves.

 

Example 1:

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

Example 2:

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

 

Constraints:

  • The number of nodes in the tree is in the range [1, 104].
  • 1 <= Node.val <= 100

Solutions

DFS or BFS.

  • /**
 * 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.
# 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.
 * 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
    }
}

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