# 129. Sum Root to Leaf Numbers

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 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:

``````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.
``````

``````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);
}
};
``````

``````class Solution {
public:
int sumNumbers(TreeNode* root) {
if (!root) return 0;
int res = 0;
stack<TreeNode*> st{{root}};
while (!st.empty()) {
TreeNode *t = st.top(); st.pop();
if (!t->left && !t->right) {
res += t->val;
}
if (t->right) {
t->right->val += t->val * 10;
st.push(t->right);
}
if (t->left) {
t->left->val += t->val * 10;
st.push(t->left);
}
}
return res;
}
};
``````

Github 同步地址：

https://github.com/grandyang/leetcode/issues/129

Path Sum

Binary Tree Maximum Path Sum

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

https://leetcode.com/problems/sum-root-to-leaf-numbers/discuss/41367/Non-recursive-preorder-traverse-Java-solution

https://leetcode.com/problems/sum-root-to-leaf-numbers/discuss/41452/Iterative-C%2B%2B-solution-using-stack-(similar-to-postorder-traversal)

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