Question

113. 路径总和 II

English Version

题目描述

给你二叉树的根节点 root 和一个整数目标和 targetSum ，找出所有 从根节点到叶子节点 路径总和等于给定目标和的路径。

叶子节点 是指没有子节点的节点。

 

示例 1：

输入：root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
输出：[[5,4,11,2],[5,8,4,5]]

示例 2：

输入：root = [1,2,3], targetSum = 5
输出：[]

示例 3：

输入：root = [1,2], targetSum = 0
输出：[]

 

提示：

• 树中节点总数在范围 [0, 5000] 内
• -1000 <= Node.val <= 1000
• -1000 <= targetSum <= 1000

解法

方法一：DFS

我们从根节点开始，递归遍历所有从根节点到叶子节点的路径，并记录路径和。当遍历到叶子节点时，如果此时路径和等于 targetSum，则将此路径加入答案。

时间复杂度 $O(n^2)$，其中 $n$ 是二叉树的节点数。空间复杂度 $O(n)$。

# 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.
 * 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);
  }
}

/**
 * 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 {
public:
  vector<vector<int>> pathSum(TreeNode* root, int targetSum) {
    vector<vector<int>> ans;
    vector<int> t;
    function<void(TreeNode*, int)> dfs = [&](TreeNode* root, int s) {
      if (!root) return;
      s -= root->val;
      t.emplace_back(root->val);
      if (!root->left && !root->right && s == 0) ans.emplace_back(t);
      dfs(root->left, s);
      dfs(root->right, s);
      t.pop_back();
    };
    dfs(root, targetSum);
    return ans;
  }
};

/**
 * 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 {
      ans = append(ans, slices.Clone(t))
    }
    dfs(root.Left, s)
    dfs(root.Right, s)
    t = t[:len(t)-1]
  }
  dfs(root, targetSum)
  return
}

// 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
  }
}

/**
 * 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;
};