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

## 129. Sum Root to Leaf Numbers

## 题目地址

<https://leetcode.com/problems/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.

```java
/**
 * 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

```java
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);
    }
  }

}
```