437PathSumIII.org

[#C] 437 Path Sum III - LeetCode

题目

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

思路

该题没有像 PathSum 一样限制需要从根节点到叶子节点。所以可以从任意的节点开始查找。

Java

/**
 * 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 {
    public int pathSum(TreeNode root, int sum) {
        if (root == null) return 0;
        // 先找包含当前节点的
        int res = findPath(root, sum);
        // 再找不包含当前节点的
        res += pathSum(root.left, sum);
        res += pathSum(root.right, sum);
        return res;
    }

    private int findPath(TreeNode node, int sum) {
        if (node == null) return 0;
        int res = 0;
        if (node.val == sum) res++;
        res += findPath(node.left, sum - node.val);
        res += findPath(node.right, sum - node.val);
        return res;
    }
}

CPP

/**
 * 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 sum) {
        if (root == NULL) return 0;

        int res = findPath(root, sum);
        res += pathSum(root->left, sum);
        res += pathSum(root->right, sum);
        return res;
    }
private:
    int findPath(TreeNode* node, int sum) {
        if (node == NULL) return 0;

        int res = 0;
        if (node->val == sum) res++;
        res += findPath(node->left, sum - node->val);
        res += findPath(node->right, sum - node->val);
        return res;
    }
};