Solution.java

/// 437. Path Sum III
/// https://leetcode.com/problems/path-sum-iii/description/
/// 时间复杂度: O(n), n为树的节点个数
/// 空间复杂度: O(h), h为树的高度
class Solution {

    /// Definition for a binary tree node.
    public static class TreeNode {
        int val;
        TreeNode left;
        TreeNode right;
        TreeNode(int x) { val = x; }
    }

    // 在以root为根节点的二叉树中,寻找和为sum的路径,返回这样的路径个数
    public int pathSum(TreeNode root, int sum) {

        if(root == null)
            return 0;

        return findPath(root, sum)
                + pathSum(root.left , sum)
                + pathSum(root.right , sum);
    }

    // 在以node为根节点的二叉树中,寻找包含node的路径,和为sum
    // 返回这样的路径个数
    private int findPath(TreeNode node, int num){

        if(node == null)
            return 0;

        int res = 0;
        if(node.val == num)
            res += 1;

        res += findPath(node.left , num - node.val);
        res += findPath(node.right , num - node.val);

        return res;
    }

    public static void main(String[] args) {

        // 手动创建Leetcode题页上的测试用例。
        // 当然, 有更好的更智能的创建二叉树的方式, 有兴趣的同学可以自行研究编写程序:)

        /*****************
         * 测试用例:
         *
         *       10
         *      /  \
         *     5   -3
         *    / \    \
         *   3   2   11
         *  / \   \
         * 3  -2   1
         *****************/
        TreeNode node1 = new TreeNode(3);
        TreeNode node2 = new TreeNode(-2);

        TreeNode node3 = new TreeNode(3);
        node3.left = node1;
        node3.right = node2;

        TreeNode node4 = new TreeNode(1);
        TreeNode node5 = new TreeNode(2);
        node5.right = node4;

        TreeNode node6 = new TreeNode(5);
        node6.left = node3;
        node6.right = node5;

        TreeNode node7 = new TreeNode(11);
        TreeNode node8 = new TreeNode(-3);
        node8.right = node7;

        TreeNode node9 = new TreeNode(10);
        node9.left = node6;
        node9.right = node8;

        System.out.println((new Solution()).pathSum(node9, 8));
    }
}