[LeetCode] 129. Sum Root to Leaf Numbers

You are given the root of a binary tree containing digits from 0 to 9 only.

Each root-to-leaf path in the tree represents a number.

For example, the root-to-leaf path 1 -> 2 -> 3 represents the number 123.
Return the total sum of all root-to-leaf numbers. Test cases are generated so that the answer will fit in a 32-bit integer.

A leaf node is a node with no children.

Example 1:

Input: root = [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: root = [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.

Constraints:
The number of nodes in the tree is in the range [1, 1000].
0 <= Node.val <= 9
The depth of the tree will not exceed 10.

求根到叶子节点数字之和。

给你一个二叉树的根节点 root ，树中每个节点都存放有一个 0 到 9 之间的数字。 每条从根节点到叶节点的路径都代表一个数字：

例如，从根节点到叶节点的路径 1 -> 2 -> 3 表示数字 123 。
计算从根节点到叶节点生成的 所有数字之和 。

叶节点 是指没有子节点的节点。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/sum-root-to-leaf-numbers
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思路

思路是前序遍历。记录一个变量 sum 存之前所有的加和，当遍历到当前节点的时候，sum *= 10 再加当前的节点值 cur.val。

复杂度

时间O(n)
空间O(n)

代码

Java实现

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    int res = 0;

    public int sumNumbers(TreeNode root) {
        helper(root, 0);
        return res;
    }

    private void helper(TreeNode root, int sum) {
        if (root == null) {
            return;
        }
        sum = sum * 10 + root.val;
        if (root.left == null && root.right == null) {
            res += sum;
            return;
        }
        helper(root.left, sum);
        helper(root.right, sum);
    }
}

JavaScript实现

/**
 * @param {TreeNode} root
 * @return {number}
 */
var sumNumbers = function (root) {
    if (root == null) {
        return 0;
    }
    var total = 0;
    helper(root, 0);
    return total;

    function helper(root, sum) {
        if (root == null) {
            return;
        }
        sum = sum * 10 + root.val;
        if (root.left == null && root.right == null) {
            total += sum;
            return;
        }
        helper(root.left, sum);
        helper(root.right, sum);
    }
};

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129. Sum Root to Leaf Numbers
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