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.
给一个数，判定树中有多少种路径能达到和为该数？
析：自己写一个方法判断以某个结点为起点能有多少条路径使和为sum
注意点： 有些数据要放在方法外面！！ 将问题简化！
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { // 注意，要把这两个数放在函数的外面，否则会递归时被清零 int all=0; int num =0; public int pathSum(TreeNode root, int sum) { if(root ==null) return 0; pathSum(root.left, sum); all+=getNumStartWith(root, sum); // 此处的num清零因为getNumStartWith每换到一个新结点就要清零 num=0; pathSum(root.right, sum); return all; } // 求以root为起点的构成和为sum的路径数目 public int getNumStartWith(TreeNode root, int sum){ if(root ==null) return 0; if(root.val==sum) num++; getNumStartWith(root.left, sum-root.val); getNumStartWith(root.right, sum-root.val); return num; } }