Question

Formatted question description: https://leetcode.ca/all/113.html

113	Path Sum II

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

For example:
Given the below binary tree and sum = 22,
              5
             / \
            4   8
           /   / \
          11  13  4
         /  \    / \
        7    2  5   1
return
[
   [5,4,11,2],
   [5,8,4,5]
]

@tag-tree

Algorithm

To find the path in the previous question “Path Sum,” we still use DFS search, but the data structure is more complex and requires a two-dimensional list.

Whenever a new node is found during DFS search, it must be saved. Once a path is found, the path saved as a one-dimensional list is saved to the final result two-dimensional list. If DFS searches a child node and finds that it is not a path sum, it needs to be removed from the one-dimensional list when returning to the previous node.

Code

import java.util.ArrayList;
import java.util.List;

public class Path_Sum_II {

	public static void main(String[] args) {
		Path_Sum_II out = new Path_Sum_II();
        Solution s = out.new Solution();

		TreeNode root = new TreeNode(1);
		root.left = new TreeNode(2);

		s.pathSum(root, 3);
	}

	/**
	 * Definition for a binary tree node.
	 * public class TreeNode {
	 *     int val;
	 *     TreeNode left;
	 *     TreeNode right;
	 *     TreeNode(int x) { val = x; }
	 * }
	 */

	public class Solution {

		List<List<Integer>> list = new ArrayList<>();

	    public List<List<Integer>> pathSum(TreeNode root, int sum) {

	    	if (root == null) {
	    		return list;
	    	}

	    	List<Integer> one = new ArrayList<>();

	        find(root, root, sum, one);

	        return list;
	    }

	    private void find(TreeNode root, TreeNode originalRoot, int sum, List<Integer> one) {

	    	if (root == null) {
	    		return;
	    	}

	    	if (root.val == sum && root.left == null && root.right == null) {

                one.add(root.val);
                // list.add(one);
                // @note:@memorize: still, above pass a reference. should be a completely new list
                list.add(new ArrayList(one));

                return;
	    	}

	    	one.add(root.val);

	    	find(root.left, root, sum - root.val, new ArrayList(one));
	    	find(root.right, root, sum - root.val, new ArrayList(one));
	    }
	}
}

############

/**
 * 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 List<List<Integer>> ans = new ArrayList<>();
    private List<Integer> t = new ArrayList<>();

    public List<List<Integer>> pathSum(TreeNode root, int targetSum) {
        dfs(root, targetSum);
        return ans;
    }

    private void dfs(TreeNode root, int s) {
        if (root == null) {
            return;
        }
        s -= root.val;
        t.add(root.val);
        if (root.left == null && root.right == null && s == 0) {
            ans.add(new ArrayList<>(t));
        }
        dfs(root.left, s);
        dfs(root.right, s);
        t.remove(t.size() - 1);
    }
}

// OJ: https://leetcode.com/problems/path-sum-ii/
// Time: O(N)
// Space: O(H)
class Solution {
    vector<vector<int>> ans;
    vector<int> tmp;
    void dfs(TreeNode *root, int target) {
        if (!root) return;
        tmp.push_back(root->val);
        target -= root->val;
        if (target == 0 && !root->left && !root->right) ans.push_back(tmp);
        dfs(root->left, target);
        dfs(root->right, target);
        tmp.pop_back();
    }
public:
    vector<vector<int>> pathSum(TreeNode* root, int target) {
        dfs(root, target);
        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:
    def pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]:
        def dfs(root, s):
            if root is None:
                return
            s += root.val
            t.append(root.val)
            if root.left is None and root.right is None and s == targetSum:
                ans.append(t[:])
            dfs(root.left, s)
            dfs(root.right, s)
            t.pop()

        ans = []
        t = []
        dfs(root, 0)
        return ans

############

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
  def pathSum(self, root, sum):
    """
    :type root: TreeNode
    :type sum: int
    :rtype: List[List[int]]
    """

    def dfs(root, s, path, res):
      if root:
        path.append(root.val)
        s -= root.val
        left = dfs(root.left, s, path, res)
        right = dfs(root.right, s, path, res)
        if not left and not right and s == 0:
          res.append(path + [])
        path.pop()
        return True

    res = []
    dfs(root, sum, [], res)
    return res

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func pathSum(root *TreeNode, targetSum int) (ans [][]int) {
	t := []int{}
	var dfs func(*TreeNode, int)
	dfs = func(root *TreeNode, s int) {
		if root == nil {
			return
		}
		s -= root.Val
		t = append(t, root.Val)
		if root.Left == nil && root.Right == nil && s == 0 {
			cp := make([]int, len(t))
			copy(cp, t)
			ans = append(ans, cp)
		}
		dfs(root.Left, s)
		dfs(root.Right, s)
		t = t[:len(t)-1]
	}
	dfs(root, targetSum)
	return
}

/**
 * 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 ans = [];
    const t = [];
    function dfs(root, s) {
        if (!root) return;
        s -= root.val;
        t.push(root.val);
        if (!root.left && !root.right && s == 0) ans.push([...t]);
        dfs(root.left, s);
        dfs(root.right, s);
        t.pop();
    }
    dfs(root, targetSum);
    return ans;
};

// 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 {
    fn dfs(
        root: Option<Rc<RefCell<TreeNode>>>,
        paths: &mut  Vec<i32>,
        mut target_sum: i32,
        res: &mut Vec<Vec<i32>>,
    ) {
        if let Some(node) = root {
            let mut node = node.borrow_mut();
            target_sum -= node.val;
            paths.push(node.val);
            if node.left.is_none() && node.right.is_none() {
                if target_sum == 0 {
                    res.push(paths.clone());
                }
            } else {
                if node.left.is_some() {
                    Self::dfs(node.left.take(), paths, target_sum, res);
                }
                if node.right.is_some() {
                    Self::dfs(node.right.take(), paths, target_sum, res);
                }
            }
            paths.pop();
        }
    }

    pub fn path_sum(root: Option<Rc<RefCell<TreeNode>>>, target_sum: i32) -> Vec<Vec<i32>> {
        let mut res = vec![];
        let mut paths = vec![];
        Self::dfs(root, &mut paths,  target_sum, &mut res);
        res
    }
}

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