LeetCode 437.：Path Sum III
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.
第一个想法就是遍历二叉树，找出每个节点的可能性。这个思路时间复杂度为O(n^2)。

	private int num;
    public int pathSum(TreeNode root, int sum) {
        if(root == null) return num;
        recur(root, sum, 0);
        pathSum(root.left, sum);
        pathSum(root.right, sum);
        return num;
    }
    private void recur(TreeNode root, int sum, int add) {
        if(root == null) {
            return;
        }
        add += root.val;
        if (add == sum) num++;
        recur(root.left, sum, add);
        recur(root.right, sum, add);
    }

改进下代码

	public int pathSum2(TreeNode root, int sum) {
        if(root == null) return 0;
        return recur(root, sum) + pathSum2(root.left, sum) + pathSum2(root.right, sum);
    }

    private int recur(TreeNode node, int sum) {
        if (node == null) return 0;
        return (node.val == sum ? 1: 0) + recur(node.left, sum-node.val) +
                recur(node.right, sum - node.val);
    }

考虑用额外空间来缩短时间复杂度，采用map，其key为从根节点到该节点这条路径上所有节点值的和，value记录该值在整个树中一共有几条这样的路径（都是从根节点开始）。则递归过程中符合要救的路径的个数:
count=map.get(sum-target) ? 0 : 1;

	public int pathSum3(TreeNode root, int target) {
        Map<Integer, Integer> map = new HashMap<Integer, Integer>();
        map.put(0, 1); //sum-target == 0,则应返回1，表示有一条合格路径
        return recur(root, 0, target, map);
    }
    private int recur(TreeNode node, int sum, int target, Map<Integer, Integer> map){
        if(node == null) return 0;
        sum += node.val;
        // 表示有count条到node的路径(从任意二叉树节点开始)符合要求
        int count = map.getOrDefault(sum-target,0);
        map.put(sum, map.getOrDefault(sum, 0)+1);
        count += recur(node.left, sum, target, map) + recur(node.right, sum, target, map);
        // 回溯过程要清除该节点的记录
        map.put(sum, map.get(sum)-1);
        return count;
    }