113. Path Sum II

最新推荐文章于 2025-09-09 16:53:03 发布

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最新推荐文章于 2025-09-09 16:53:03 发布

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文章标签： c# runtime

原文链接：http://www.cnblogs.com/chucklu/p/10925533.html

博客主要围绕二叉树展开，给定一个二叉树和一个特定的和值，要找出所有从根节点到叶子节点的路径，使得这些路径上节点值的和等于给定的和值，同时说明了叶子节点是没有子节点的节点，并给出示例。

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

Note: A leaf is a node with no children.

Example:

Given the below binary tree and sum = 22,

      5
     / \
    4   8
   /   / \
  11  13  4
 /  \    / \
7    2  5   1

Return:

[
   [5,4,11,2],
   [5,8,4,5]
]

 private List<IList<int>> list = new List<IList<int>>();

        public IList<IList<int>> PathSum(TreeNode root, int sum)
        {
            Chuck(root, sum, 0, string.Empty);
            return list;
        }

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

            tempSum = tempSum + node.val;
            var left = node.left;
            var right = node.right;
            if (string.IsNullOrEmpty(str))
            {
                str = $"{node.val}";
            }
            else
            {
                str = $"{str},{node.val}";
            }

            if (left == null && right == null)
            {
                if (tempSum == sum)
                {
                    var tempList = str.Split(',').Select(x => Convert.ToInt32(x)).ToList();
                    list.Add(tempList);
                }
            }
            else
            {
                Chuck(left, sum, tempSum, str);
                Chuck(right, sum, tempSum, str);
            }
        }