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Formatted question description: https://leetcode.ca/all/1022.html

# 1022. Sum of Root To Leaf Binary Numbers (Easy)

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.

Companies:
Amazon

Related Topics:
Tree

## Solution 1.

• /**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int sumRootToLeaf(TreeNode root) {
int sum = 0;
nodeQueue.offer(root);
numQueue.offer(root.val);
while (!nodeQueue.isEmpty()) {
TreeNode node = nodeQueue.poll();
int num = numQueue.poll();
TreeNode left = node.left, right = node.right;
if (left == null && right == null)
sum += num;
else {
if (left != null) {
nodeQueue.offer(left);
numQueue.offer(num * 2 + left.val);
}
if (right != null) {
nodeQueue.offer(right);
numQueue.offer(num * 2 + right.val);
}
}
}
return sum;
}
}

############

/**
* 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 {
public int sumRootToLeaf(TreeNode root) {
return dfs(root, 0);
}

private int dfs(TreeNode root, int t) {
if (root == null) {
return 0;
}
t = (t << 1) | root.val;
if (root.left == null && root.right == null) {
return t;
}
return dfs(root.left, t) + dfs(root.right, t);
}
}

• // OJ: https://leetcode.com/problems/sum-of-root-to-leaf-binary-numbers/
// Time: O(N)
// Space: O(H)
class Solution {
public:
int sumRootToLeaf(TreeNode* root, int pre = 0) {
if (!root) return 0;
int val = (pre << 1) + root->val;
if (!root->left && !root->right) return val;
return sumRootToLeaf(root->left, val) + sumRootToLeaf(root->right, val);
}
};

• # Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def sumRootToLeaf(self, root: TreeNode) -> int:
def dfs(root, t):
if root is None:
return 0
t = (t << 1) | root.val
if root.left is None and root.right is None:
return t
return dfs(root.left, t) + dfs(root.right, t)

return dfs(root, 0)

############

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
def sumRootToLeaf(self, root):
"""
:type root: TreeNode
:rtype: int
"""
if not root: return 0
self.res = 0
self.dfs(root, root.val)
return self.res

def dfs(self, root, preSum):
if not root.left and not root.right:
self.res = (self.res + preSum) % (10 ** 9 + 7)
return
if root.left:
self.dfs(root.left, preSum * 2 + root.left.val)
if root.right:
self.dfs(root.right, preSum * 2 + root.right.val)

• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func sumRootToLeaf(root *TreeNode) int {
var dfs func(root *TreeNode, t int) int
dfs = func(root *TreeNode, t int) int {
if root == nil {
return 0
}
t = (t << 1) | root.Val
if root.Left == nil && root.Right == nil {
return t
}
return dfs(root.Left, t) + dfs(root.Right, t)
}

return dfs(root, 0)
}

• /**
* Definition for a binary tree node.
* class TreeNode {
*     val: number
*     left: TreeNode | null
*     right: TreeNode | null
*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.left = (left===undefined ? null : left)
*         this.right = (right===undefined ? null : right)
*     }
* }
*/

function sumRootToLeaf(root: TreeNode | null): number {
const dfs = (root: TreeNode | null, num: number) => {
if (root == null) {
return 0;
}
const { val, left, right } = root;
num = (num << 1) | val;
if (left == null && right == null) {
return num;
}
return dfs(left, num) + dfs(right, num);
};
return dfs(root, 0);
}


• // Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, mut num: i32) -> i32 {
if root.is_none() {
return 0;
}
let root = root.as_ref().unwrap().borrow();
num = (num << 1) | root.val;
if root.left.is_none() && root.right.is_none() {
return num;
}
Self::dfs(&root.left, num) + Self::dfs(&root.right, num)
}

pub fn sum_root_to_leaf(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
Self::dfs(&root, 0)
}
}