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中等
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第 16 场双周赛 Q3
树
深度优先搜索
广度优先搜索
二叉树

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

我们可以使用广度优先搜索，逐层遍历二叉树，并在遍历到每一层时计算该层的节点值之和。遍历完成后，返回最后一层的节点值之和。

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。

Python3

# 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)):
                node = q.popleft()
                ans += node.val
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
        return ans

Java

/**
 * 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 k = q.size(); k > 0; --k) {
                TreeNode node = q.poll();
                ans += node.val;
                if (node.left != null) {
                    q.offer(node.left);
                }
                if (node.right != null) {
                    q.offer(node.right);
                }
            }
        }
        return ans;
    }
}

C++

/**
 * 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 k = q.size(); k; --k) {
                TreeNode* node = q.front();
                q.pop();
                ans += node->val;
                if (node->left) {
                    q.push(node->left);
                }
                if (node->right) {
                    q.push(node->right);
                }
            }
        }
        return ans;
    }
};

Go

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

TypeScript

/**
 * 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 q: TreeNode[] = [root];
    let ans = 0;
    while (q.length) {
        const nq: TreeNode[] = [];
        ans = 0;
        for (const { val, left, right } of q) {
            ans += val;
            left && nq.push(left);
            right && nq.push(right);
        }
        q = nq;
    }
    return ans;
}

Rust

// 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;
use std::collections::VecDeque;

impl Solution {
    pub fn deepest_leaves_sum(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
        let mut q = VecDeque::new();
        q.push_back(root);
        let mut ans = 0;
        while !q.is_empty() {
            ans = 0;
            for _ in 0..q.len() {
                if let Some(Some(node)) = q.pop_front() {
                    let node = node.borrow();
                    ans += node.val;
                    if node.left.is_some() {
                        q.push_back(node.left.clone());
                    }
                    if node.right.is_some() {
                        q.push_back(node.right.clone());
                    }
                }
            }
        }
        ans
    }
}

方法二：DFS

我们可以使用深度优先搜索，递归遍历二叉树，并在遍历的过程中记录当前节点的深度，以及最大深度和最深叶子节点的和。遍历到当前节点时，如果当前节点的深度等于最大深度，则将当前节点的值加到最深叶子节点的和中；如果当前节点的深度大于最大深度，则将最大深度更新为当前节点的深度，并将最深叶子节点的和更新为当前节点的值。

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。

Python3

# 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

Java

/**
 * 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);
    }
}

C++

/**
 * 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 mx = 0, ans = 0;
        auto dfs = [&](auto&& dfs, TreeNode* root, int i) {
            if (!root) {
                return;
            }
            if (i == mx) {
                ans += root->val;
            } else if (i > mx) {
                mx = i;
                ans = root->val;
            }
            dfs(dfs, root->left, i + 1);
            dfs(dfs, root->right, i + 1);
        };
        dfs(dfs, root, 1);
        return ans;
    }
};

Go

/**
 * 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
}

TypeScript

/**
 * 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 [ans, mx] = [0, 0];
    const dfs = (root: TreeNode | null, i: number) => {
        if (!root) {
            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);
    };
    dfs(root, 1);
    return ans;
}

C

/**
 * 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;
}