Description:
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:
- 5 -> 3
- 5 -> 2 -> 1
- -3 -> 11
题目描述:
给定一个二叉树，它的每个结点都存放着一个整数值。
找出路径和等于给定数值的路径总数。
路径不需要从根节点开始，也不需要在叶子节点结束，但是路径方向必须是向下的（只能从父节点到子节点）。
二叉树不超过1000个节点，且节点数值范围是 [-1000000,1000000] 的整数。
示例：
root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8
10
/ \
5 -3
/ \ \
3 2 11
/ \ \
3 -2 1
返回 3。和等于 8 的路径有:
- 5 -> 3
- 5 -> 2 -> 1
- -3 -> 11
思路:
- 深度优先搜索和层次遍历结合
- 双重递归
时间复杂度O(n ^ 2), 空间复杂度O(n)
代码:
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;
queue q;
q.push(root);
while (!q.empty()) {
int s = q.size();
for (int i = 0; i < s; i++) {
TreeNode* cur = q.front();
if (cur -> left) q.push(cur -> left);
if (cur -> right) q.push(cur -> right);
dfs(cur, sum);
q.pop();
}
}
return result;
}
private:
int result = 0;
void dfs(TreeNode* root, int sum) {
if (!root) return;
sum -= root -> val;
if (!sum) result++;
dfs(root -> left, sum);
dfs(root -> right, sum);
}
};
Java:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
private int result = 0;
private void dfs(TreeNode root, int sum) {
if (root == null) return;
sum -= root.val;
if (sum == 0) result++;
dfs(root.left, sum);
dfs(root.right, sum);
}
public int pathSum(TreeNode root, int sum) {
if (root == null) return 0;
List list = new ArrayList<>();
list.add(root);
while (!list.isEmpty()) {
int s = list.size();
for (int i = 0; i < s; i++) {
TreeNode cur = list.get(0);
if (cur.left != null) list.add(cur.left);
if (cur.right != null) list.add(cur.right);
dfs(cur, sum);
list.remove(0);
}
}
return result;
}
}
Python:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def pathSum(self, root: TreeNode, sum: int) -> int:
if not root:
return 0
return self.check(root, sum) + self.pathSum(root.left, sum) + self.pathSum(root.right, sum)
def check(self, root: TreeNode, sum: int) -> int:
if not root:
return 0
result = 0
if sum == root.val:
result += 1
result += self.check(root.left, sum - root.val) + self.check(root.right, sum- root.val)
return result