[LeetCode] Path Sum

题目

原题地址

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.

For example:
Given the below binary tree and sum = 22,

          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

递归查找，每个节点判断一下是否叶子节点，如果是则判断value之和是否与sum相同；如果不是叶子节点，就递归查找它的左右子节点。

python代码

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

class Solution:
    def hasPathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: bool
        """
        if root == None:
            return False
        if root.left == None and root.right == None:
            # leaf node
            return root.val == sum
        return self.hasPathSum(root.left, sum-root.val) or self.hasPathSum(root.right, sum-root.val)

Path Sum

进阶版Path Sum，不仅判断是否存在这样的路径，而是要求给出所有满足要求的路径。
原题地址

思路

总体思想与上一题类似，但是需要增加一个参数list来保存路径，所以需要重新写一个递归函数。注意往list里append值之后要对应的pop，否则后续对list的操作也会反映到最终结果上，因为最终返回的是一个List[List[int]]，其中的每个list是浅拷贝。

python代码

# 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 search(self, root, sum, ls):
        if root == None:
            return
        if root.left == None and root.right == None:
            # leaf node
            if sum == root.val:
                ls.append(root.val)
                self.result.append(ls[:])
                ls.pop()
        else:
            # not leaf node
            ls.append(root.val)
            self.search(root.left, sum-root.val, ls)
            self.search(root.right, sum-root.val, ls)
            ls.pop()

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