# [LeetCode] 129. Sum Root to Leaf Numbers 求根到叶节点数字之和

2021年09月15日 阅读数：1

Given a binary tree containing digits from `0-9` only, each root-to-leaf path could represent a number.html

An example is the root-to-leaf path `1->2->3` which represents the number `123`.java

Find the total sum of all root-to-leaf numbers.node

Note: A leaf is a node with no children.python

Example:git

```Input: [1,2,3]
1
/ \
2   3
Output: 25
Explanation:
The root-to-leaf path `1->2` represents the number `12`.
The root-to-leaf path `1->3` represents the number `13`.
Therefore, sum = 12 + 13 = `25`.```

Example 2:app

```Input: [4,9,0,5,1]
4
/ \
9   0
/ \
5   1
Output: 1026
Explanation:
The root-to-leaf path `4->9->5` represents the number 495.
The root-to-leaf path `4->9->1` represents the number 491.
The root-to-leaf path `4->0` represents the number 40.
Therefore, sum = 495 + 491 + 40 = `1026`.```

Java:htm

```public int sumNumbers(TreeNode root) {
return sum(root, 0);
}

public int sum(TreeNode n, int s){
if (n == null) return 0;
if (n.right == null && n.left == null) return s*10 + n.val;
return sum(n.left, s*10 + n.val) + sum(n.right, s*10 + n.val);
}
```

Python:

```# Time:  O(n)
# Space: O(h), h is height of binary tree
class TreeNode(object):
def __init__(self, x):
self.val = x
self.left = None
self.right = None

class Solution(object):
# @param root, a tree node
# @return an integer
def sumNumbers(self, root):
return self.sumNumbersRecu(root, 0)

def sumNumbersRecu(self, root, num):
if root is None:
return 0

if root.left is None and root.right is None:
return num * 10 + root.val

return self.sumNumbersRecu(root.left, num * 10 + root.val) + self.sumNumbersRecu(root.right, num * 10 + root.val)　```

Python:  bfs + stack

```def sumNumbers1(self, root):
if not root:
return 0
stack, res = [(root, root.val)], 0
while stack:
node, value = stack.pop()
if node:
if not node.left and not node.right:
res += value
if node.right:
stack.append((node.right, value*10+node.right.val))
if node.left:
stack.append((node.left, value*10+node.left.val))
return res
```

Python:  bfs + queue

```#
def sumNumbers2(self, root):
if not root:
return 0
queue, res = collections.deque([(root, root.val)]), 0
while queue:
node, value = queue.popleft()
if node:
if not node.left and not node.right:
res += value
if node.left:
queue.append((node.left, value*10+node.left.val))
if node.right:
queue.append((node.right, value*10+node.right.val))
return res
```

Python:  Recursive

```def sumNumbers(self, root):
self.res = 0
self.dfs(root, 0)
return self.res

def dfs(self, root, value):
if root:
#if not root.left and not root.right:
#    self.res += value*10 + root.val
self.dfs(root.left, value*10+root.val)
#if not root.left and not root.right:
#    self.res += value*10 + root.val
self.dfs(root.right, value*10+root.val)
if not root.left and not root.right:
self.res += value*10 + root.val　```

C++:

```/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int sumNumbers(TreeNode *root) {
return sumNumbersDFS(root, 0);
}
int sumNumbersDFS(TreeNode *root, int sum) {
if (!root) return 0;
sum = sum * 10 + root->val;
if (!root->left && !root->right) return sum;
return sumNumbersDFS(root->left, sum) + sumNumbersDFS(root->right, sum);
}
};
```

[LeetCode] 112. Path Sum 路径和

[LeetCode] 124. Binary Tree Maximum Path Sum 求二叉树的最大路径和