leetcode -- Sum Root to Leaf Numbers -- dfs

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https://leetcode.com/problems/sum-root-to-leaf-numbers/

思路1

用dfs preorder求就行

class Solution(object):    def dfs(self, root, subres, res):        if root:            if root.left == None and root.right == None:                #print subres + str(root.val)                res[0] += int(subres + str(root.val))            else:                self.dfs(root.left, subres + str(root.val), res)                self.dfs(root.right, subres + str(root.val), res)    def sumNumbers(self, root):        """        :type root: TreeNode        :rtype: int        """        res = [0]        self.dfs(root, '', res)        return res[0]

思路2

http://www.cnblogs.com/zuoyuan/p/3721420.html还提供了另一种递归的办法

class Solution:    # @param root, a tree node    # @return an integer    def sum(self, root, preSum):#preSum记录root的父节点到global root的Sum。        if root==None: return 0        preSum = preSum*10 + root.val#这里跟位操作有点像，preSum向左移动一位，然后加上root.val。        if root.left==None and root.right==None: return preSum        return self.sum(root.left, preSum)+self.sum(root.right, preSum)    def sumNumbers(self, root):        return self.sum(root, 0)

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