Level:
Easy
题目描述:
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 1Return 3. The paths that sum to 8 are:1. 5 -> 32. 5 -> 2 -> 13. -3 -> 11
思路分析:
由于题目中所说的路径,不局限于从根到叶子节点,任何一个节点都可以作为路径的开始节点和终止节点,所以我们以根节点作为开始节点,查找和为sum的路径数,然后分别以根的左孩子和右孩子为起始节点去查找和为sum的路径数,依次递归向下推导,得到最终的结果。
代码:
/**public class TreeNode{ int vla; TreeNode left; TreeNode right; public TreeNode(int val){ this.val=val; }}*/public class Soulution{ public int pathSum(TreeNode root,int sum){ if(root==null) return 0; int res=0; res=pathCheck(root,sum); res=res+pathSum(root.left,sum); res=res+pathSum(root.right,sum); return res; } public int pathCheck(TreeNode root,int sum){ if(root==null) return 0; int count=0; if(sum==root.val)//当sum等于root.val时证明存在一条路径和为sum count++; count=count+pathCheck(root.left,sum-root.val); count=count+pathCheck(root.right,sum-root.val); return count; }}