Path Sum

地址:点击打开链接

这道题是实际是遍历树然后求和，只不过加了一个参数，所以要会写BFS和DFS，以及层次遍历

# 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
        """
        return self.DFS(0, sum, root)
    def DFS(self, curSum, sum, root):
        if root == None:
            return False
        if root.left == None and root.right == None:
            return (root.val + curSum) == sum
        return self.DFS(curSum + root.val, sum, root.left) or self.DFS(curSum + root.val, sum, root.right)

自己写了一个深度遍历，不知道对不对

def DFSTree(self, root, nodeList):
        if root == None:
            return None
        if root.left == None and root.right == None:
            nodeList.append(root.val)
        else:
            nodeList.extend(self.DFSTree(root.left))
            nodeList.extend(self.DFSTree(root.right))
        return nodeList