1022. Sum of Root To Leaf Binary Numbers (E)

Sum of Root To Leaf Binary Numbers (E)

题目

Given a binary tree, each node has value 0 or 1. Each root-to-leaf path represents a binary number starting with the most significant bit. For example, if the path is 0 -> 1 -> 1 -> 0 -> 1, then this could represent 01101 in binary, which is 13.

For all leaves in the tree, consider the numbers represented by the path from the root to that leaf.

Return the sum of these numbers.

Example 1:

Input: [1,0,1,0,1,0,1]
Output: 22
Explanation: (100) + (101) + (110) + (111) = 4 + 5 + 6 + 7 = 22

Note:

1. The number of nodes in the tree is between 1 and 1000.
2. node.val is 0 or 1.
3. The answer will not exceed 2^31 - 1.

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题意

给定一个二叉树，从根结点到叶结点的路径构成一个二进制数，求所有二进制数的和。

思路

直接dfs处理即可。

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代码实现

Java

class Solution {
    public int sumRootToLeaf(TreeNode root) {
        return dfs(root, 0);
    }

    private int dfs(TreeNode root, int sum) {
        sum = sum * 2 + root.val;
        if (root.left == null && root.right == null) {
            return sum;
        } else if (root.left == null) {
            return dfs(root.right, sum);
        } else if (root.right == null) {
            return dfs(root.left, sum);
        } else {
            return dfs(root.left, sum) + dfs(root.right, sum);
        }
    }
}