[LeetCode] 129. Sum Root to Leaf Numbers_Medium tag: DFS
2018-08-08 00:40 Johnson_强生仔仔 阅读(281) 评论(0) 编辑 收藏 举报Given a binary tree containing digits from 0-9
only, each root-to-leaf path could represent a number.
An example is the root-to-leaf path 1->2->3
which represents the number 123
.
Find the total sum of all root-to-leaf numbers.
Note: A leaf is a node with no children.
Example:
Input: [1,2,3] 1 / \ 2 3 Output: 25 Explanation: The root-to-leaf path1->2
represents the number12
. The root-to-leaf path1->3
represents the number13
. Therefore, sum = 12 + 13 =25
.
Example 2:
Input: [4,9,0,5,1] 4 / \ 9 0 / \ 5 1 Output: 1026 Explanation: The root-to-leaf path4->9->5
represents the number 495. The root-to-leaf path4->9->1
represents the number 491. The root-to-leaf path4->0
represents the number 40. Therefore, sum = 495 + 491 + 40 =1026
.
思路就是DFS, 然后将node 和当前的path组成的数字string一起append进入stack, 判断如果是leaf, 将num转换为int加入在ans中.
1. Constraints
1)None => 0
2. Ideas
DFS T: O(n) S; O(n)
3. Code
class Solution: def sumRootLeaf(self, root): if not root: return 0 stack, ans = [(root, str(root.val))], 0 while stack: node, num = stack.pop() if not node.right and not node.left: ans += int(num) if node.left: stack.append((node.left, num + str(node.left.val))) if node.right: stack.append((node.right, num + str(node.right.val))) return ans