題目:
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題目鏈接
題意:
給一棵二叉樹,其中每個節點包含一個整數值。給定一個值sum,要求找到路徑和爲sum的路徑的個數。
路徑不需要從root開始或結束,但必須向下(只能從父節點到子節點)
數的節點數不超過1000個,取值範圍在-1000000到1000000之間。
由於路徑並不要求必須以葉子爲終點,所以不能寫從葉子向上遞加的算法,通過觀察得出,每一個非葉子結點都可以作爲一段路徑的起點,而每一個非根節點結點都可以作爲一段路經的終點,所以我們可以遍歷整個樹,枚舉所有可能的起點,對與每一個起點進行搜索,以其開始的路徑是否有長度爲sum的,有的話,ans++,當所有遍歷結束,返回ans。
代碼如下:
class Solution {
public:
int target, ans = 0;
void upToSum(TreeNode* node, int sum) {
if (!node) return;
sum += node->val;
if (sum == target) {
ans++;
}
upToSum(node->left, sum);
upToSum(node->right, sum);
}
void dfs(TreeNode * node) {
if (!node) return;
upToSum(node, 0);
dfs(node->left);
dfs(node->right);
}
int pathSum(TreeNode* root, int sum) {
target = sum;
dfs(root);
return ans;
}
};