二叉树的路径和（没必要以根节点为起始）Path Sum III

问题：

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

解决：

① 这道题就是给了一个二叉树和一个目标和sum，找出全部路径，这个路径的和等于sum，只容许从父节点到子节点的路线。使用DFS（Dpeth-first Search，深度优先搜索）查找知足条件的路径。

/**  *

 Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {//30ms
    public int pathSum(TreeNode root, int sum) {
        if(root == null) //必须判断，不然会越界
            return 0;
        return dfs(root, sum) + pathSum(root.left, sum) + pathSum(root.right, sum);
    }
    private int dfs(TreeNode root, int sum){
        int count = 0;
        if(root == null)
            return count;
        if(sum == root.val)
            count ++;
        count += dfs(root.left,sum - root.val);
        count += dfs(root.right,sum - root.val);
        return count;
    }
}

② 用map记录当前遍历路径中的子路径权值和对应出现的次数。

1. sum表示从根节点到某节点X，路径的权值和，则遍历至X节点时，当前的路径和curSum刚好与sum相等，此时count = map.getOrDefault(curSum - target,0) = map.get(0) = 1;

2. 若sum为某段子路径的权值和，如：x1->x2->x3->x4......中sum等于节点x3与节点x4的权值和，即sum = sum（x3+x4）。则遍历至x2时， m[curSum]++; 处已经记录了m[curSum] = m[sum（x1+x2）] = 1,遍历至x4时curSum = sum（x1+x2+x3+x4）,则res = m[curSum - sum] = m[sum（x1+x2+x3+x4） - sum（x3+x4）] = m[sum（x1+x2）] = 1。

public class Solution {//26ms     public int pathSum(TreeNode root, int sum) {         Map<Integer,Integer> map = new HashMap<Integer,Integer>();//map(curSum，个数)         map.put(0,1);//初始权值为0，个数为1         return dfs(root,sum,map,0);        }     public int dfs(TreeNode root,int sum,Map<Integer,Integer> map,int curSum){         if(root == null) return 0;         curSum += root.val;         int count = map.getOrDefault(curSum - sum,0);//获取当前子节点包含权值为 sum的子段的个数         map.put(curSum,map.getOrDefault(curSum,0) + 1);//记录当前子节点权值的个数加1         count += dfs(root.left,sum,map,curSum) + dfs(root.right,sum,map,curSum);         map.put(curSum,map.get(curSum) - 1);         return count;        } }