leetcode112

/**
 * 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 = 0;
                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)

 

java版本实现：

class Solution {
    public boolean hasPathSum(TreeNode root, int targetSum) {
        if(root == null){
            return false;
        }
        if(root.left == null && root.right == null && root.val == targetSum){
            return true;
        }else{
            return hasPathSum(root.left,targetSum-root.val) || hasPathSum(root.right,targetSum-root.val);
        }
    }
}