129-Sum Root to Leaf Numbers

Description

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.

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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.

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問題描述

給定二叉樹， 其節點的值僅為0-9間的數字， 每個從根節點到葉子節點的路徑可以代表一個數字。

例如根節點到葉子節點的路徑為1->2->3, 代表123.

返回所有根節點到葉子節點組成的數之和

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問題分析

從根節點到葉子節點的遞迴過程中注意將之前求得的和進位即可

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解法1

public class Solution {
    public int sumNumbers(TreeNode root) {
        if(root == null) return 0;

        int total = 0;
        Queue<TreeNode> q = new LinkedList<>();
        Queue<Integer> sumq = new LinkedList<>();    
        q.offer(root);
        sumq.offer(root.val);

        while(!q.isEmpty()){
            TreeNode current = q.poll();
            int partialSum = sumq.poll();
            if(current.left == null && current.right == null){
                total += partialSum;
            }else{
                if(current.left != null){
                    q.offer(current.left);
                    sumq.offer(partialSum * 10 + current.left.val);
                }
                if(current.right !=null){
                    q.offer(current.right);
                    sumq.offer(partialSum * 10 + current.right.val);
                }
            }

        }

        return total;
    }
}

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解法2

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

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