113. Path Sum II

Problem Description

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 equals targetSum. 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.

Examples

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: []

# 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
def pathSum(root: Optional[TreeNode], targetSum: int) -> List[List[int]]:
    def dfs(node: Optional[TreeNode], remaining: int, path: List[int], result: List[List[int]]) -> None:
        if not node:
            return
        
        # Add current node to path
        path.append(node.val)
        
        # Check if it's a leaf node and sum matches
        if not node.left and not node.right and remaining == node.val:
            result.append(path[:])  # Make a copy of the path
        
        # Recursively check left and right subtrees
        dfs(node.left, remaining - node.val, path, result)
        dfs(node.right, remaining - node.val, path, result)
        
        # Backtrack by removing current node
        path.pop()
    
    result = []
    dfs(root, targetSum, [], result)
    return result

/**
 * 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 void dfs(TreeNode node, int remaining, List path, List> result) {
        if (node == null) {
            return;
        }
        
        // Add current node to path
        path.add(node.val);
        
        // Check if it's a leaf node and sum matches
        if (node.left == null && node.right == null && remaining == node.val) {
            result.add(new ArrayList<>(path));  // Make a copy of the path
        }
        
        // Recursively check left and right subtrees
        dfs(node.left, remaining - node.val, path, result);
        dfs(node.right, remaining - node.val, path, result);
        
        // Backtrack by removing current node
        path.remove(path.size() - 1);
    }
    
    public List> pathSum(TreeNode root, int targetSum) {
        List> result = new ArrayList<>();
        dfs(root, targetSum, new ArrayList<>(), result);
        return result;
    }
}

/**
 * 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:
    void dfs(TreeNode* node, int remaining, vector& path, vector>& result) {
        if (!node) {
            return;
        }
        
        // Add current node to path
        path.push_back(node->val);
        
        // Check if it's a leaf node and sum matches
        if (!node->left && !node->right && remaining == node->val) {
            result.push_back(path);  // Vector will be copied
        }
        
        // Recursively check left and right subtrees
        dfs(node->left, remaining - node->val, path, result);
        dfs(node->right, remaining - node->val, path, result);
        
        // Backtrack by removing current node
        path.pop_back();
    }
    
public:
    vector> pathSum(TreeNode* root, int targetSum) {
        vector> result;
        vector path;
        dfs(root, targetSum, path, result);
        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) {
    const dfs = (node, remaining, path, result) => {
        if (!node) {
            return;
        }
        
        // Add current node to path
        path.push(node.val);
        
        // Check if it's a leaf node and sum matches
        if (!node.left && !node.right && remaining === node.val) {
            result.push([...path]);  // Make a copy of the path
        }
        
        // Recursively check left and right subtrees
        dfs(node.left, remaining - node.val, path, result);
        dfs(node.right, remaining - node.val, path, result);
        
        // Backtrack by removing current node
        path.pop();
    };
    
    const result = [];
    dfs(root, targetSum, [], result);
    return result;
};

/**
 * 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 {
    private void Dfs(TreeNode node, int remaining, IList path, IList> result) {
        if (node == null) {
            return;
        }
        
        // Add current node to path
        path.Add(node.val);
        
        // Check if it's a leaf node and sum matches
        if (node.left == null && node.right == null && remaining == node.val) {
            result.Add(new List(path));  // Make a copy of the path
        }
        
        // Recursively check left and right subtrees
        Dfs(node.left, remaining - node.val, path, result);
        Dfs(node.right, remaining - node.val, path, result);
        
        // Backtrack by removing current node
        path.RemoveAt(path.Count - 1);
    }
    
    public IList> PathSum(TreeNode root, int targetSum) {
        IList> result = new List>();
        Dfs(root, targetSum, new List(), result);
        return result;
    }
}