求一個二叉樹所有從根節點到葉子結點構成的字符串轉為數字之后的數值之和，簡單DFS/BFS
Python版——DFS
1 #129
2 #Runtime: 20 ms (Beats 46.49%)
3 #Memory: 13.4 MB (Beats 79.53%)
4
5 # Definition for a binary tree node.
6 # class TreeNode(object):
7 # def __init__(self, val=0, left=None, right=None):
8 # self.val = val
9 # self.left = left
10 # self.right = right
11 class Solution(object):
12 def sumNumbers(self, root):
13 """
14 :type root: TreeNode
15 :rtype: int
16 """
17 self.ans = 0
18 def DFS(node, t):
19 if not node:
20 return
21 if node.left or node.right:
22 DFS(node.left, t * 10 + int(node.val))
23 DFS(node.right, t * 10 + int(node.val))
24 else:
25 self.ans += t * 10 + int(node.val)
26 return
27
28 DFS(root, 0)
29 return self.ans
C++版——BFS
1 //129
2 //Runtime: 24 ms (Beats 19.15%)
3
4 /**
5 * Definition for binary tree
6 * struct TreeNode {
7 * int val;
8 * TreeNode *left;
9 * TreeNode *right;
10 * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
11 * };
12 */
13 class Solution {
14 public:
15 struct Que {
16 int s;
17 TreeNode *p;
18 }q[100010];
19
20 int sumNumbers(TreeNode *root) {
21 int l = 0, r = 1, res = 0;
22 if(root == NULL) return 0;
23 q[0].p = root;
24 q[0].s = root->val;
25 while(l < r) {
26 if(q[l].p->left == NULL && q[l].p->right == NULL) {
27 res += q[l].s;
28 }
29 if(q[l].p->left != NULL) {
30 q[r].p = q[l].p->left;
31 q[r].s = q[l].s * 10 + q[r].p->val;
32 ++r;
33 }
34 if(q[l].p->right != NULL) {
35 q[r].p = q[l].p->right;
36 q[r].s = q[l].s * 10 + q[r].p->val;
37 ++r;
38 }
39 ++l;
40 }
41 return res;
42 }
43 };