leetcode解题 : Path Sum III (437)

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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

解法1：DFS / Recursion

这题考察DFS的基本知识。因为考虑的是每一个从上到下的path，那么应该要想到要用DFS。
对于每一个节点，如果包括自己，则可以递归运算left 和right，而要match的数则变成了sum - val。
如果不包括自己，则直接运算left和right，最后将两个情况相加就是结果。
实际上这里就是用了一个preorder遍历，回想一下preorder的算法：
visit root;
visit left;
visit right;
这里也一样：
visit root => compute number of paths with root (dfs(root, sum))
visit left => compute number of paths with left (pathSum(root->left, sum))
visit right => compute number of paths with right (pathSum(root->right, sum))

C++

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/**
 * 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 == NULL) {
            return 0;
        }
        int left = pathSum(root->left, sum);
        int right = pathSum(root->right, sum);
        return dfs(root, sum) + left + right;
    }
    int dfs(TreeNode* root, int sum) {
        int res = 0;
        if (root == NULL) {
            return res;
        }
        if (root->val == sum) {
            res++;
        }
        res += dfs(root->left, sum - root->val);
        res += dfs(root->right, sum - root->val);
        return res;
    }
};

Java

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/**
 * 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;
        }
        int left = pathSum(root.left, sum);
        int right = pathSum(root.right, sum);
        // Count the number of paths starting from the current root
        int startingFromThisRoot = dfs(root, sum);
        return startingFromThisRoot + left + right;
    }
    private int dfs(TreeNode root, int sum) {
        int res = 0;
        if (root == null) return res;
        if (root.val == sum) ++res;
        res += dfs(root.left, sum - root.val);
        res += dfs(root.right, sum - root.val);
        return res;
    }
}