Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

sum = 22

1
2
3
4
5
6
7
      5
     / \
    4   8
   /   / \
  11  13  4
 /  \      \
7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

思路

这一题也是利用深度优先遍历(DFS)的递归方法来进行查找。

需要注意的是，只有自根节点到叶子节点的路径才算是一条符合题意的路径，其它情况都不需要额外考虑了。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
# 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 (root):
            return self.DFS(root, sum)
        else:
            return False
        
    def DFS(self, root, sum):
        if (root):
            val = root.val
            sum -= val
            if(root.left is None and root.right is None):
                if (sum == 0):
                    return True
                else:
                    return False
            else:
                return self.DFS(root.left, sum) or self.DFS(root.right, sum)
        else:
            return False