# LeetCode.1022.Sum of Root To Leaf Binary Numbers 从根到叶的二进制数之和

## 题目描述

1022 Sum of Root To Leaf Binary Numbers
https://leetcode-cn.com/problems/sum-of-root-to-leaf-binary-numbers/

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]

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.

## 解题过程

import javafx.util.Pair;
private static class SolutionV2020 {
public int sumRootToLeaf(TreeNode root) {
if (null == root) {
return 0;
}
int sum = 0;
// pair的右值是树结点的parent所表示的二进制数和
Deque<Pair<TreeNode, Integer>> queue = new LinkedList<>();
queue.offer(new Pair<>(root, 0));
while (!queue.isEmpty()) {
Pair<TreeNode, Integer> pair = queue.poll();
TreeNode node = pair.getKey();
// 当前节点node表示的二进制数和
int thisSum = pair.getValue() * 2 + node.val;
// node是叶子节点，累加和到sum
if (null == node.left && null == node.right) {
sum += thisSum;
}
if (null != node.left) {
queue.offer(new Pair<>(node.left, thisSum));
}
if (null != node.right) {
queue.offer(new Pair<>(node.right, thisSum));
}
}
return sum;
}
}

## GitHub代码

algorithms/leetcode/leetcode/_1022_SumOfRootToLeafBinaryNumbers.java
https://github.com/masikkk/algorithms/blob/master/leetcode/leetcode/_1022_SumOfRootToLeafBinaryNumbers.java

436

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