129. Sum Root to Leaf Numbers-优快云博客

本文介绍了一种算法，用于计算二叉树中所有从根节点到叶子节点路径所表示的数字之和。通过递归深度优先搜索(DFS)策略，文章详细解释了如何遍历树结构并累加路径值，最终得到总和。

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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.

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.

Problem URL

Solution

给一棵二叉树，从根到叶子的一条路径表示一个数。返回这些数的加和。类似于Path Sum。递归DFS解决。

Using a global variable sum to store the result. Then for each path, calculate the value it denotes. When come to the leaf node, add the value to sum. else recursively visit left child and right child.

Code

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private int sum = 0;
    public int sumNumbers(TreeNode root) {
        if (root == null){
            return 0;
        }
        helper(root, 0);
        return sum;
    }
    
    private void helper(TreeNode node, int curValue){
        if (node == null){
            return;
        }
        curValue = curValue * 10 + node.val;
        if (node.left == null && node.right == null){
            sum += curValue;          
        }
        else{
            helper(node.left, curValue);
            helper(node.right, curValue);
        }
    }
}