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
从任意一点dfs, 和dfs return 当前, 左节点, 右节点的模板
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public int pathSum(TreeNode root, int sum) {
if(root==null) return 0;
//helper(root,sum) 当前节点开始
//pathSum(root.left,sum) 当前节点左节点开始
//pathSum(root.right,sum) 当前节点右节点开始
return helper(root,sum)+pathSum(root.left,sum)+pathSum(root.right,sum);
}
private int helper(TreeNode root,int sum){
if(root==null) return 0;
int count=0;
if(root.val==sum) count++;
return count+helper(root.left,sum-root.val)+helper(root.right,sum-root.val);
}
}
原文：http://www.cnblogs.com/apanda009/p/7233620.html