题目如下：
Given a binary tree, each node has value
0or1. Each root-to-leaf path represents a binary number starting with the most significant bit. For example, if the path is0 -> 1 -> 1 -> 0 -> 1, then this could represent01101in binary, which is13.For all leaves in the tree, consider the numbers represented by the path from the root to that leaf.
Return the sum of these numbers.
Example 1:
Input: [1,0,1,0,1,0,1] Output: 22 Explanation: (100) + (101) + (110) + (111) = 4 + 5 + 6 + 7 = 22
Note:
- The number of nodes in the tree is between
1and1000.- node.val is
0or1.- The answer will not exceed
2^31 - 1.
解题思路：没啥好说的，递归遍历树吧。
代码如下：
# 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): res = 0 def recursive(self,node,path): path += str(node.val) if node.left == None and node.right == None: self.res += int(path,2) return if node.left != None: self.recursive(node.left,path) if node.right != None: self.recursive(node.right, path) def sumRootToLeaf(self, root): """ :type root: TreeNode :rtype: int """ self.res = 0 self.recursive(root,‘‘) return self.res % (10 ** 9+ 7)
【leetcode】1022. Sum of Root To Leaf Binary Numbers
原文：https://www.cnblogs.com/seyjs/p/10673718.html
