[LeetCode] 437. Path Sum III

轉自LeetCode

You are given a binary tree in which each node contains an integer value.

Find the number of paths that sum to a given value.

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

The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.

Example:

root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8

      10
     /  \
    5   -3
   / \    \
  3   2   11
 / \   \
3  -2   1

Return 3. The paths that sum to 8 are:

1.  5 -> 3
2.  5 -> 2 -> 1
3. -3 -> 11

<Solution>

想法如下(參考資料)

• 用一個 hash map 記錄從 root 到前一個 node 的總合，以及出現的次數

• 如果 curSum - target 在 hash map 裡面有資料，代表在前面一定有總合為 target 的 path  存在

• 利用 DFS 的方式來歷遍

code 如下

C++

	/**
	* Definition for a binary tree node.
	* struct TreeNode {
	* int val;
	* TreeNode *left;
	* TreeNode *right;
	* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
	* };
	*/
	class Solution {
	public:
	int pathSum(TreeNode* root, int sum) {
	if(!root) {
	return 0;
	}
	unordered_map<int, int> preSumMap;
	preSumMap[0] = 1;
	return dfs(root, 0, sum, preSumMap);
	}
	int dfs(TreeNode *node, int curSum, const int &target, unordered_map<int, int> &map) {
	if(!node) {
	return 0;
	}
	curSum += node->val;
	const int key = curSum - target;
	int cnt = map.count(key) ? map[key] : 0;
	map[curSum]++;
	cnt += dfs(node->left, curSum, target, map) + dfs(node->right, curSum, target, map);
	map[curSum]--;
	return cnt;
	}
	};

view raw pathSum3.cpp hosted with ❤ by GitHub

另一種解法純粹是 recursive，不用 hashmap

想法如下

• 第一層 recursive，選 root，或者選 root.left，或者選 root.right

• 第二層 recursive，純粹歷遍，找有沒有答案

code 如下

Java(參考解法)

	/**
	* Definition for a binary tree node.
	* public class TreeNode {
	* int val;
	* TreeNode left;
	* TreeNode right;
	* TreeNode(int x) { val = x; }
	* }
	*/
	class Solution {
	public int pathSum(TreeNode root, int sum) {
	if(root == null) {
	return 0;
	}
	return dfs(root, sum) + pathSum(root.left, sum) + pathSum(root.right, sum);
	}
	private int dfs(TreeNode root, int sum) {
	if(root == null) {
	return 0;
	}
	return (root.val == sum ? 1 : 0) + dfs(root.left, sum - root.val) + dfs(root.right, sum - root.val);
	}
	}

view raw pathSum3.java hosted with ❤ by GitHub