路径总和

给你二叉树的根节点 root 和一个表示目标和的整数 targetSum ，判断该树中是否存在 根节点到叶子节点 的路径，这条路径上所有节点值相加等于目标和 targetSum 。

叶子节点 是指没有子节点的节点。

示例 1：

输入：root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
输出：true

示例 2：

输入：root = [1,2,3], targetSum = 5
输出：false

示例 3：

输入：root = [1,2], targetSum = 0
输出：false

提示：

• 树中节点的数目在范围 [0, 5000] 内
• -1000 <= Node.val <= 1000
• -1000 <= targetSum <= 1000

示例代码1：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def hasPathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: bool
        """
        if not root:
            return False
        queue = collections.deque([root])
        queue_val = collections.deque([root.val])

        while queue:
            node = queue.popleft()
            tmp = queue_val.popleft()
            if (not node.left) and (not node.right):
                if tmp == sum:
                    return True
                continue
            if node.left:
                queue.append(node.left)
                queue_val.append(node.left.val + tmp)
            if node.right:
                queue.append(node.right)
                queue_val.append(node.right.val + tmp)
        return False

示例代码2：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def hasPathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: bool
        """
        if not root:
            return False
        if (not root.left) and (not root.right):
            return sum == root.val
        return self.hasPathSum(root.left, sum-root.val) or self.hasPathSum(root.right, sum-root.val)

示例代码3（栈）：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def hasPathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: bool
        """
        if not root:
            return False
        stack = []
        stack.append((root, root.val))
        while stack:
            node, path = stack.pop(0)
            if (not node.left) and (not node.right) and path == sum:
                return True
            if node.left:
                stack.append((node.left, node.left.val+path))
            if node.right:
                stack.append((node.right, node.right.val+path))
        return False