/**
* Definition for a binary tree node.
* public class TreeNode {
* public int val;
* public TreeNode left;
* public TreeNode right;
* public TreeNode(int x) { val = x; }
* }
*/
public class Solution
{
Stack<TreeNode> S = new Stack<TreeNode>();
List<List<TreeNode>> list = new List<List<TreeNode>>();
private void postNode(TreeNode node)
{
if (node != null)
{
S.Push(node); if (node.left != null)
{
postNode(node.left);
} if (node.right != null)
{
postNode(node.right);
} if (node.left == null && node.right == null)
{
list.Add(S.ToList());
}
S.Pop();
}
} public bool HasPathSum(TreeNode root, int sum)
{
postNode(root); foreach (var l in list)
{
var psum = ;
foreach (var d in l)
{
psum += d.val;
}
if (psum == sum)
{
return true;
}
} return false;
}
}
https://leetcode.com/problems/path-sum/#/description
补充一个python的实现：
class Solution:
def hasPathSum(self, root: TreeNode, sum: int) -> bool:
if root == None:
return False
if root.left == None and root.right == None and root.val == sum:
return True
return self.hasPathSum(root.left,sum-root.val) or self.hasPathSum(root.right,sum-root.val)