113. Path Sum II

Problem Description

Given the root of a binary tree and an integer targetSum, return all root-to-leaf paths where each path’s sum equals targetSum.

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]]

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

Solution

DFS (Backtracking)

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 pathSum(self, root: TreeNode, targetSum: int) -> List[List[int]]:
        self.paths = []
        self._dfs(root, targetSum, [])
        return self.paths


    def _dfs(self, root, target, curr):
        if not root:
            return

        if not root.left and not root.right and root.val == target:
            self.paths.append(curr[:] + [root.val])
            return

        if root.left:
            self._dfs(root.left, target - root.val, curr + [root.val])

        if root.right:
            self._dfs(root.right, target - root.val, curr + [root.val])

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 List<List<Integer>> pathSum(TreeNode root, int targetSum) {
        List<List<Integer>> ret = new ArrayList<List<Integer>>();
        if (root == null) {
            return ret;
        }

        List<Integer> curr = new ArrayList<Integer>();
        curr.add(root.val);
        dfs(root, targetSum - root.val, curr, ret);

        return ret;
    }


    private void dfs(TreeNode root, int target, List<Integer> curr, List<List<Integer>> ret) {

        if (root.left == null && root.right == null && target == 0) {
            ret.add(new ArrayList(curr));
            return;
        }

        if (root.left == null && root.right == null) {
            return;
        }

        if (root.left != null) {
            curr.add(root.left.val);
            dfs(root.left, target - root.left.val, curr, ret);
            curr.remove(curr.size() - 1);
        }

        if (root.right != null) {
            curr.add(root.right.val);
            dfs(root.right, target - root.right.val, curr, ret);
            curr.remove(curr.size() - 1);
        }
    }

}

Complexity Analysis

• Time Complexity
  • O(n)
• Space Complexity
  • O(n^2)