路径总和(Go)

给定一个二叉树和一个目标和,找到所有从根节点到叶子节点路径总和等于给定目标和的路径。

说明: 叶子节点是指没有子节点的节点。

func pathSum(root *TreeNode, sum int) [][]int {
	ret := [][]int{}
	path := []int{}
	dfs(&ret, root, path, sum)
	return ret
}

func dfs(ret *[][]int, root *TreeNode, path []int, target int) {
	switch {
	case root == nil:
		return
	case root.Left == nil && root.Right == nil && root.Val == target:
		dst := make([]int, len(path)+1)
		copy(dst, append(path, root.Val))
		*ret = append(*ret, dst)
		return
	}
	path = append(path, root.Val)
	dfs(ret, root.Left, path, target-root.Val)
	dfs(ret, root.Right, path, target-root.Val)
}

  

原文地址:https://www.cnblogs.com/lvpengbo/p/14452642.html