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2235. 两整数相加：
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 。
样例 1：
输入：
root = [1,2,3,4,5,null,6,7,null,null,null,null,8]
输出：
15
样例 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, ] 之间。
- 1 <= Node.val <= 100
分析
- 面对这道算法题目，二当家陷入了沉思。
- 最直观的方式就是先遍历一次计算树中的最深层数，然后再遍历统计节点和，但是这样需要遍历两次，有没有办法仅遍历一次呢？
题解
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 {
private int maxDep = 0;
private int ans = 0;
public int deepestLeavesSum(TreeNode root) {
dfs(root, 1);
return ans;
}
private void dfs(TreeNode root, int dep) {
if (root == null) {
return;
}
if (dep > maxDep) {
maxDep = dep;
ans = root.val;
} else if (dep == maxDep) {
ans += root.val;
}
dfs(root.left, dep + 1);
dfs(root.right, dep + 1);
}
}
c
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
void dfs(struct TreeNode* root, int dep, int* maxDep, int* ans) {
if (!root) {
return;
}
if (dep > *maxDep) {
*maxDep = dep;
*ans = root->val;
} else if (dep == *maxDep) {
*ans += root->val;
}
dfs(root->left, dep + 1, maxDep, ans);
dfs(root->right, dep + 1, maxDep, ans);
}
int deepestLeavesSum(struct TreeNode* root){
int maxDep = 0;
int ans = 0;
dfs(root, 1, &maxDep, &ans);
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 {
private:
int maxDep = 0;
int ans = 0;
void dfs(TreeNode* root, int dep) {
if (!root) {
return;
}
if (dep > maxDep) {
maxDep = dep;
ans = root->val;
} else if (dep == maxDep) {
ans += root->val;
}
dfs(root->left, dep + 1);
dfs(root->right, dep + 1);
}
public:
int deepestLeavesSum(TreeNode* root) {
dfs(root, 1);
return ans;
}
};
python
# 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 __init__(self):
self.ans = 0
self.max_dep = 0
def deepestLeavesSum(self, root: Optional[TreeNode]) -> int:
def dfs(node: Optional[TreeNode], dep: int) -> None:
if not node:
return
if dep > self.max_dep:
self.max_dep = dep
self.ans = node.val
elif dep == self.max_dep:
self.ans += node.val
dfs(node.left, dep + 1)
dfs(node.right, dep + 1)
dfs(root, 1)
return self.ans
go
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func deepestLeavesSum(root *TreeNode) int {
ans := 0
maxDep := 0
var dfs func(node *TreeNode, dep int)
dfs = func(node *TreeNode, dep int) {
if node == nil {
return
}
if dep > maxDep {
maxDep = dep
ans = node.Val
} else if dep == maxDep {
ans += node.Val
}
dfs(node.Left, dep+1)
dfs(node.Right, dep+1)
}
dfs(root, 1)
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;
impl Solution {
pub fn deepest_leaves_sum(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
fn dfs(node: Option<Rc<RefCell<TreeNode>>>, dep: i32, maxDep: &mut i32, ans: &mut i32) {
if let Some(node) = node {
if dep > *maxDep {
*maxDep = dep;
*ans = node.borrow().val;
} else if dep == *maxDep {
*ans += node.borrow().val;
}
dfs(node.borrow().left.clone(), dep + 1, maxDep, ans);
dfs(node.borrow().right.clone(), dep + 1, maxDep, ans);
}
}
let mut ans = 0;
let mut maxDep = 0;
dfs(root, 1, &mut maxDep, &mut ans);
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 = 0;
let maxDep = 0;
function dfs(node: TreeNode | null, dep: number): void {
if (!node) {
return
}
if (dep > maxDep) {
maxDep = dep;
ans = node.val;
} else if(dep == maxDep) {
ans += node.val;
}
dfs(node.left, dep+1);
dfs(node.right, dep+1);
}
dfs(root, 1);
return ans;
};
原题传送门：https://leetcode-cn.com/problems/deepest-leaves-sum/
非常感谢你阅读本文~
放弃不难，但坚持一定很酷~
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