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# 129.Sum-Root-to-Leaf-Numbers

## 题目描述

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.

## 代码

### Approach #1 Iterative Preorder Traversal

Complexity Analysis
• Time complexity: O(N) since one has to visit each node.
• Space complexity: up to O(H) to keep the stack, where H is a tree height.
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int sumNumbers(TreeNode root) {
int rootToLeaf = 0, currNumber = 0;
Deque<Pair<TreeNode, Integer>> stack = new ArrayDeque();
stack.push(new Pair(root, 0));
while (!stack.isEmpty()) {
Pair<TreeNode, Integer> p = stack.pop();
root = p.getKey();
currNumber = p.getValue();
if (root != null) {
currNumber = currNumber * 10 + root.val;
if (root.left == null && root.right == null) {
rootToLeaf += currNumber;
} else {
stack.push(new Pair(root.right, currNumber));
stack.push(new Pair(root.left, currNumber));
}
}
}
return rootToleaf;
}
}

### Approach #2 Recursive Preorder Traversal

class Solution {
int rootToLeaf = 0;
public int sumNumbers(TreeNode root) {
preorder(root, 0);
return rootToLeaf;
}
public void preorder(TreeNode r, int currNumber) {
if (r != null) {
currNumber = currNumber * 10 + r.val;
if (r.left == null && r.right == null) {
rootToLeaf += currNumber;
}
preorder(r.left, currNumber);
preorder(r.right, currNumber);
}
}
}