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Author: @wkw, @lonyehan, @radojicic23 | https://leetcode.com/problems/path-sum-ii/ |
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https://leetcode.com/problems/path-sum-ii/
Given the root of a binary tree and an integer targetSum, return all root-to-leaf paths where the sum of the node values in the path equalstargetSum. Each path should be returned as a list of the node values, not node references.
A root-to-leaf path is a path starting from the root and ending at any leaf node. A leaf is a node with no children.
Example 1:
Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
Output: [[5,4,11,2],[5,8,4,5]]
Explanation: There are two paths whose sum equals targetSum:
5 + 4 + 11 + 2 = 22
5 + 8 + 4 + 5 = 22
Example 2:
Input: root = [1,2,3], targetSum = 5
Output: []
Example 3:
Input: root = [1,2], targetSum = 0
Output: []
Constraints:
- The number of nodes in the tree is in the range
[0, 5000]. -1000 <= Node.val <= 1000-1000 <= targetSum <= 1000
class Solution {
public:
vector<vector<int>> ans;
// the idea is to use dfs to traverse the tree from all root to leaf paths
// `path` is used to store the current route
// `remainingSum` is used to store thre remaining sum that we need with the initial value `targetSum`.
// we substract it from the node value when we traverse down the tree
// if we arrive the leaf and the the remaining sum is eqaul to leaf node value
// then we can add `path` to ans
// e.g. path = [5,4,11,2] and remainingSum = targetSum = 22
// traverse node 5, remainingSum = 22 - 5 = 17
// traverse node 4, remainingSum = 17 - 4 = 13
// traverse node 11, remainingSum = 13 - 11 = 2
// traverse node 2, remainingSum = 2 - 2 = 0
// remainingSum is 0 which means the sum of current path is eqaul to targetSum
// so how to find another path?
// we can backtrack here
// we can pop back a node and try another
// e.g. path = [5, 4, 11, 7]
// pop 7 out = [5, 4, 11]
// push 2 = [5, 4, 11, 2]
// ... etc
void dfs(TreeNode* node, vector<int>& path, int remainingSum) {
// if it is nullptr, then return
if (!node) return;
// add the current node val to path
path.push_back(node-> val);
// !node->left && !node->right : check if it is a leaf node
// remainingSum == node->val: check if the remaining sum - node->val == 0
// if both condition is true, then we find one of the paths
if (!node->left && !node->right && remainingSum == node->val) ans.push_back(path);
// traverse left sub tree with updated remaining sum
// we don't need to check if left sub tree is nullptr or not
// as we handle it in the first line of this function
dfs(node-> left, path, remainingSum - node-> val);
// traverse right sub tree with updated remaining sum
// we don't need to check if right sub tree is nullptr or not
// as we handle it in the first line of this function
dfs(node-> right, path, remainingSum - node-> val);
// backtrack
path.pop_back();
}
vector<vector<int>> pathSum(TreeNode* root, int targetSum) {
// used to store current route
vector<int> path;
// dfs from the root
dfs(root, path, targetSum);
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:
# the idea is to use dfs to traverse the tree from all root to leaf paths
# `path` is used to store the current route
# `remainingSum` is used to store thre remaining sum that we need with the initial value `targetSum`.
# we substract it from the node value when we traverse down the tree
# if we arrive the leaf and the the remaining sum is eqaul to leaf node value
# then we can add `path` to ans
# e.g. path = [5,4,11,2] and remainingSum = targetSum = 22
# traverse node 5, remainingSum = 22 - 5 = 17
# traverse node 4, remainingSum = 17 - 4 = 13
# traverse node 11, remainingSum = 13 - 11 = 2
# traverse node 2, remainingSum = 2 - 2 = 0
# remainingSum is 0 which means the sum of current path is eqaul to targetSum
# so how to find another path?
# we can backtrack here
# we can pop back a node and try another
# e.g. path = [5, 4, 11, 7]
# pop 7 out = [5, 4, 11]
# push 2 = [5, 4, 11, 2]
# ... etc
def dfs(self, root, path, ans, remainingSum):
# if it is None, then return
if not root:
return
# add the current node val to path
path.append(root.val)
# !node.left && !node.right : check if it is a leaf node
# remainingSum == node.val: check if the remaining sum - node.val == 0
# if both condition is true, then we find one of the paths
if not root.left and not root.right and remainingSum == root.val:
# lists passed a function are just references (i.e. Pass By Reference)
# and path.pop() would pop them all eventually
# therefore you need to create a new list there
# or your can use ans.append(path[:]) / ans.append(path.copy())
ans.append(list(path))
# traverse left sub tree with updated remaining sum
# we don't need to check if left sub tree is nullptr or not
# as we handle it in the first line of this function
self.dfs(root.left, path, ans, remainingSum - root.val)
# traverse right sub tree with updated remaining sum
# we don't need to check if right sub tree is nullptr or not
# as we handle it in the first line of this function
self.dfs(root.right, path, ans, remainingSum - root.val)
# backtrack
path.pop()
def pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]:
ans = []
self.dfs(root, [], ans, targetSum)
return ans/**
* 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 {
List<List<Integer>> ans = new ArrayList<List<Integer>>();
// the idea is to use dfs to traverse the tree from all root to leaf paths
// `path` is used to store the current route
// `remainingSum` is used to store thre remaining sum that we need with the initial value `targetSum`.
// we substract it from the node value when we traverse down the tree
// if we arrive the leaf and the the remaining sum is eqaul to leaf node value
// then we can add `path` to ans
// e.g. path = [5,4,11,2] and remainingSum = targetSum = 22
// traverse node 5, remainingSum = 22 - 5 = 17
// traverse node 4, remainingSum = 17 - 4 = 13
// traverse node 11, remainingSum = 13 - 11 = 2
// traverse node 2, remainingSum = 2 - 2 = 0
// remainingSum is 0 which means the sum of current path is eqaul to targetSum
// so how to find another path?
// we can backtrack here
// we can pop back a node and try another
// e.g. path = [5, 4, 11, 7]
// pop 7 out = [5, 4, 11]
// push 2 = [5, 4, 11, 2]
// ... etc
private void dfs(TreeNode node, List<Integer> path, int remainingSum) {
// if it is null, then return
if (node == null) return;
// add the current node val to path
path.add(node.val);
// !node.left && !node.right : check if it is a leaf node
// remainingSum == node.val: check if the remaining sum - node.val == 0
// if both condition is true, then we find one of the paths
if (node.left == null && node.right == null && remainingSum == node.val) ans.add(new ArrayList<>(path));
// traverse left sub tree with updated remaining sum
// we don't need to check if left sub tree is nullptr or not
// as we handle it in the first line of this function
this.dfs(node.left, path, remainingSum - node.val);
// traverse right sub tree with updated remaining sum
// we don't need to check if right sub tree is nullptr or not
// as we handle it in the first line of this function
this.dfs(node.right, path, remainingSum - node.val);
// backtrack
path.remove(path.size() - 1);
}
public List<List<Integer>> pathSum(TreeNode root, int targetSum) {
List<Integer> path = new ArrayList<Integer>();
dfs(root, path, targetSum);
return ans;
}
}/**
* Definition for a binary tree node.
* public class TreeNode {
* public int val;
* public TreeNode left;
* public TreeNode right;
* public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
public class Solution {
public List<IList<int>> result;
public void PreOrder(TreeNode root, int targetSum, List<int> list, int sum = 0) {
if(root == null) return;
sum += root.val;
list.Add(root.val);
// Is this node is leaf and sum equal to target?
if(sum == targetSum && root.left == null && root.right == null) {
result.Add(list);
}
PreOrder(root.left, targetSum, new List<int>(list), sum);
PreOrder(root.right, targetSum, new List<int>(list), sum);
}
public IList<IList<int>> PathSum(TreeNode root, int targetSum) {
result = new List<IList<int>>();
PreOrder(root, targetSum, new List<int>());
return result;
}
}/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @param {number} targetSum
* @return {number[][]}
*/
var pathSum = function (root, targetSum) {
function dfs(root, remainingSum, path, ans) {
if (!root) return ans;
path.push(root.val);
if (!root.left && !root.right && root.val === remainingSum) {
ans.push(path.slice());
}
dfs(root.left, remainingSum - root.val, path, ans);
dfs(root.right, remainingSum - root.val, path, ans);
path.pop();
return ans;
}
return dfs(root, targetSum, [], []);
};