LeetCode Path Sum II

LeetCode解題之Path Sum II

原題

找出一棵二叉樹所有的從根節點到某一葉子節點的路徑，該路徑上所有節點的和為一個特定值。

注意點：

例子:

輸入:

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

輸出:

[
   [5,4,11,2],
   [5,8,4,5]
]

解題思路

Path Sum 是判斷是否有這樣一條路徑，現在要把所有的路徑都求出來，那只要在dfs時將符合要求的路徑加入到結果集。注意加入結果集的數據不要是引用，否則可能之後會再次被修改。

AC源碼

# 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 pathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: List[List[int]]
        """
        result = []
        self._pathSum(root, sum, [], result)
        return result

    def _pathSum(self, root, sum, curr, result):
        if not root:
            return
        sum -= root.val
        if sum == 0 and root.left is None and root.right is None:
            result.append(curr + [root.val])
        if root.left:
            self._pathSum(root.left, sum, curr + [root.val], result)
        if root.right:
            self._pathSum(root.right, sum, curr + [root.val], result)


if __name__ == "__main__":
    None