437. Path Sum III

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原文链接：http://www.cnblogs.com/chucklu/p/10925750.html

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博客围绕二叉树展开，每个节点包含整数值，需找出路径和等于给定值的路径数量。路径无需从根节点或叶子节点开始或结束，但必须向下遍历，树节点不超1000个，节点值范围在 -1000000 到 1000000 之间。

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

 private int _count;
        private int _target;

        public int PathSum(TreeNode root, int sum)
        {
            _target = sum;
            Iterate(root);
            return _count;
        }

        private void Iterate(TreeNode node)
        {

            Chuck(node, 0);
            if (node?.left == null && node?.right == null)
            {
                return;
            }
            Iterate(node.left);
            Iterate(node.right);
        }

        private void Chuck(TreeNode node, int sum)
        {
            if (node == null)
            {
                return;
            }

            sum = sum + node.val;
            if (sum == _target)
            {
                _count++;
            }
            if (node.left == null && node.right == null)
            {
                return;
            }
            Chuck(node.left, sum);
            Chuck(node.right, sum);
        }