| comments | difficulty | edit_url |
|---|---|---|
true |
Medium |
You are given a binary tree in which each node contains an integer value (which might be positive or negative). Design an algorithm to count the number of paths that sum to a given value. The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).
Example:
Given the following tree and sum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ / \
7 2 5 1
Output:
3 Explanation: Paths that have sum 22 are: [5,4,11,2], [5,8,4,5], [4,11,7]
Note:
node number <= 10000
We can use the idea of prefix sum to recursively traverse the binary tree, and use a hash table
We design a recursive function
The recursive process of the function
- If the current node
$node$ is null, return$0$ . - Calculate the prefix sum
$s$ on the path from the root node to the current node. - Use
$cnt[s - sum]$ to represent the number of paths with the path sum equal to$sum$ and the path ends at the current node, where$cnt[s - sum]$ is the count of the prefix sum equal to$s - sum$ in$cnt$ . - Add the count of the prefix sum
$s$ by$1$ , i.e.,$cnt[s] = cnt[s] + 1$ . - Recursively traverse the left and right child nodes of the current node, i.e., call the functions
$dfs(node.left, s)$ and$dfs(node.right, s)$ , and add their return values. - After the return value is calculated, subtract the count of the prefix sum
$s$ of the current node by$1$ , i.e., execute$cnt[s] = cnt[s] - 1$ . - Finally, return the answer.
The time complexity is
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def pathSum(self, root: TreeNode, sum: int) -> int:
def dfs(root: TreeNode, s: int):
if root is None:
return 0
s += root.val
ans = cnt[s - sum]
cnt[s] += 1
ans += dfs(root.left, s)
ans += dfs(root.right, s)
cnt[s] -= 1
return ans
cnt = Counter({0: 1})
return dfs(root, 0)/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
private Map<Long, Integer> cnt = new HashMap<>();
private int target;
public int pathSum(TreeNode root, int sum) {
cnt.put(0L, 1);
target = sum;
return dfs(root, 0);
}
private int dfs(TreeNode root, long s) {
if (root == null) {
return 0;
}
s += root.val;
int ans = cnt.getOrDefault(s - target, 0);
cnt.merge(s, 1, Integer::sum);
ans += dfs(root.left, s);
ans += dfs(root.right, s);
cnt.merge(s, -1, Integer::sum);
return ans;
}
}/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int pathSum(TreeNode* root, int sum) {
unordered_map<long long, int> cnt;
cnt[0] = 1;
function<int(TreeNode*, long long)> dfs = [&](TreeNode* root, long long s) {
if (!root) {
return 0;
}
s += root->val;
int ans = cnt[s - sum];
++cnt[s];
ans += dfs(root->left, s);
ans += dfs(root->right, s);
--cnt[s];
return ans;
};
return dfs(root, 0);
}
};/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func pathSum(root *TreeNode, sum int) int {
cnt := map[int]int{0: 1}
var dfs func(*TreeNode, int) int
dfs = func(root *TreeNode, s int) int {
if root == nil {
return 0
}
s += root.Val
ans := cnt[s-sum]
cnt[s]++
ans += dfs(root.Left, s)
ans += dfs(root.Right, s)
cnt[s]--
return ans
}
return dfs(root, 0)
}/**
* 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 pathSum(root: TreeNode | null, sum: number): number {
const cnt: Map<number, number> = new Map();
cnt.set(0, 1);
const dfs = (root: TreeNode | null, s: number): number => {
if (!root) {
return 0;
}
s += root.val;
let ans = cnt.get(s - sum) ?? 0;
cnt.set(s, (cnt.get(s) ?? 0) + 1);
ans += dfs(root.left, s);
ans += dfs(root.right, s);
cnt.set(s, (cnt.get(s) ?? 0) - 1);
return ans;
};
return dfs(root, 0);
}// 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::cell::RefCell;
use std::collections::HashMap;
use std::rc::Rc;
impl Solution {
pub fn path_sum(root: Option<Rc<RefCell<TreeNode>>>, sum: i32) -> i32 {
let mut cnt = HashMap::new();
cnt.insert(0, 1);
return Self::dfs(root, sum, 0, &mut cnt);
}
fn dfs(
root: Option<Rc<RefCell<TreeNode>>>,
sum: i32,
s: i32,
cnt: &mut HashMap<i32, i32>,
) -> i32 {
if let Some(node) = root {
let node = node.borrow();
let s = s + node.val;
let mut ans = *cnt.get(&(s - sum)).unwrap_or(&0);
*cnt.entry(s).or_insert(0) += 1;
ans += Self::dfs(node.left.clone(), sum, s, cnt);
ans += Self::dfs(node.right.clone(), sum, s, cnt);
*cnt.entry(s).or_insert(0) -= 1;
return ans;
}
return 0;
}
}/* class TreeNode {
* var val: Int
* var left: TreeNode?
* var right: TreeNode?
*
* init(_ val: Int, _ left: TreeNode? = nil, _ right: TreeNode? = nil) {
* self.val = val
* self.left = left
* self.right = right
* }
* }
*/
class Solution {
private var cnt: [Int: Int] = [:]
private var target: Int = 0
func pathSum(_ root: TreeNode?, _ sum: Int) -> Int {
cnt[0] = 1
target = sum
return dfs(root, 0)
}
private func dfs(_ root: TreeNode?, _ s: Int) -> Int {
guard let root = root else {
return 0
}
let newSum = s + root.val
let ans = cnt[newSum - target, default: 0]
cnt[newSum, default: 0] += 1
let leftPaths = dfs(root.left, newSum)
let rightPaths = dfs(root.right, newSum)
cnt[newSum, default: 0] -= 1
return ans + leftPaths + rightPaths
}
}