Question

437. Path Sum III

中文文档

Description

Given the root of a binary tree and an integer targetSum, return _the number of paths where the sum of the values along the path equals_ targetSum.

The path does not need to start or end at the root or a leaf, but it must go downwards (i.e., traveling only from parent nodes to child nodes).

 

Example 1:

Input: root = [10,5,-3,3,2,null,11,3,-2,null,1], targetSum = 8
Output: 3
Explanation: The paths that sum to 8 are shown.

Example 2:

Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
Output: 3

 

Constraints:

• The number of nodes in the tree is in the range [0, 1000].
• -109 <= Node.val <= 109
• -1000 <= targetSum <= 1000

Solutions

Solution 1

# 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) -> int:
    def dfs(node, s):
      if node is None:
        return 0
      s += node.val
      ans = cnt[s - targetSum]
      cnt[s] += 1
      ans += dfs(node.left, s)
      ans += dfs(node.right, s)
      cnt[s] -= 1
      return ans

    cnt = Counter({0: 1})
    return dfs(root, 0)

/**
 * 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 Map<Long, Integer> cnt = new HashMap<>();
  private int targetSum;

  public int pathSum(TreeNode root, int targetSum) {
    cnt.put(0L, 1);
    this.targetSum = targetSum;
    return dfs(root, 0);
  }

  private int dfs(TreeNode node, long s) {
    if (node == null) {
      return 0;
    }
    s += node.val;
    int ans = cnt.getOrDefault(s - targetSum, 0);
    cnt.merge(s, 1, Integer::sum);
    ans += dfs(node.left, s);
    ans += dfs(node.right, s);
    cnt.merge(s, -1, Integer::sum);
    return ans;
  }
}

/**
 * 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:
  int pathSum(TreeNode* root, int targetSum) {
    unordered_map<long, int> cnt;
    cnt[0] = 1;
    function<int(TreeNode*, long)> dfs = [&](TreeNode* node, long s) -> int {
      if (!node) return 0;
      s += node->val;
      int ans = cnt[s - targetSum];
      ++cnt[s];
      ans += dfs(node->left, s) + dfs(node->right, s);
      --cnt[s];
      return ans;
    };
    return dfs(root, 0);
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func pathSum(root *TreeNode, targetSum int) int {
  cnt := map[int]int{0: 1}
  var dfs func(*TreeNode, int) int
  dfs = func(node *TreeNode, s int) int {
    if node == nil {
      return 0
    }
    s += node.Val
    ans := cnt[s-targetSum]
    cnt[s]++
    ans += dfs(node.Left, s) + dfs(node.Right, s)
    cnt[s]--
    return ans
  }
  return dfs(root, 0)
}

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *   val: number
 *   left: TreeNode | null
 *   right: TreeNode | null
 *   constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 *   }
 * }
 */

function pathSum(root: TreeNode | null, targetSum: number): number {
  const cnt: Map<number, number> = new Map();
  const dfs = (node: TreeNode | null, s: number): number => {
    if (!node) {
      return 0;
    }
    s += node.val;
    let ans = cnt.get(s - targetSum) ?? 0;
    cnt.set(s, (cnt.get(s) ?? 0) + 1);
    ans += dfs(node.left, s);
    ans += dfs(node.right, s);
    cnt.set(s, (cnt.get(s) ?? 0) - 1);
    return ans;
  };
  cnt.set(0, 1);
  return dfs(root, 0);
}