[LeetCode130]Sum Root to Leaf Numbers-优快云博客

本文链接：https://blog.youkuaiyun.com/sbitswc/article/details/34554797

本文探讨了如何通过遍历二叉树来计算从根节点到叶子节点的所有路径所代表的数字之和。通过深度优先搜索算法，我们可以逐层累加路径上的数字，并在到达叶子节点时将总和加入最终结果。

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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.

For example,

    1
   / \
  2   3

The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.

Return the sum = 12 + 13 = 25.

Analysis:

Once we see this kind of problem, no matter what sum is required to output, "all root-to-leaf" phrase reminds us the classic Tree Traversal or Depth-First-Search algorithm. Then according to the specific problem, compute and store the values we need. Here in this problem, while searching deeper, add the values up (times 10 + current value), and add the sum to final result if meet the leaf node (left and right child are both NULL).

Java

public class Solution {
    int result;
	public int sumNumbers(TreeNode root) {
        result = 0;
        getSum(root, 0);
        return result;
    }
	public void getSum(TreeNode root, int path){
		if(root==null) return;
		path = path*10+root.val;
		if(root.left==null && root.right==null){
			result+=path;
			return;
		}
		getSum(root.left, path);
		getSum(root.right, path);
	}
}

c++

void sumSubpath(TreeNode* root, int &sum, int path){
    if(root == NULL) return;
    path = path*10 + root->val;
    if(root->left == NULL && root->right == NULL){
        sum += path;
        return;
    }
    sumSubpath(root->left, sum, path);
    sumSubpath(root->right, sum, path);
}
int sumNumbers(TreeNode *root) {
    int sum = 0;
    int path = 0;
    sumSubpath(root,sum,0);
    return sum;
}