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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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