Sum of Root To Leaf Binary Numbers - LeetCode
You are given the root of a binary tree where each node has a 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. The test cases are generated so that the answer fits in a 32-bits integer.
Example 1:
Input: root = [1,0,1,0,1,0,1] Output: 22 Explanation: (100) + (101) + (110) + (111) = 4 + 5 + 6 + 7 = 22
Example 2:
Input: root = [0] Output: 0
Constraints:
The number of nodes in the tree is in the range [1, 1000].
Node.val is 0 or 1.
- code
class Solution:
def sumRootToLeaf(self, root: Optional[TreeNode]) -> int:
self.res = 0
def dfs(root, path):
if not root.left and not root.right:
self.res += int("".join(path + [str(root.val)]), 2)
if root.left:
dfs(root.left, path + [str(root.val)])
if root.right:
dfs(root.right, path + [str(root.val)])
dfs(root, [])
return self.res
- code
class Solution:
def sumRootToLeaf(self, root: Optional[TreeNode]) -> int:
self.res = 0
def dfs(root, val):
if root:
cur = (val << 1) + root.val
if not root.left and not root.right:
self.res += cur
dfs(root.left, cur)
dfs(root.right, cur)
dfs(root, 0)
return self.res
- code
class Solution {
public int sumRootToLeaf(TreeNode root) {
int rootToLeaf = 0, currNumber = 0;
Deque<Pair<TreeNode, Integer>> stack = new ArrayDeque();
stack.push(new Pair(root, 0));
while (!stack.isEmpty()) {
Pair<TreeNode, Integer> p = stack.pop();
root = p.getKey();
currNumber = p.getValue();
if (root != null) {
currNumber = (currNumber << 1) | root.val;
// if it's a leaf, update root-to-leaf sum
if (root.left == null && root.right == null) {
rootToLeaf += currNumber;
} else {
stack.push(new Pair(root.right, currNumber));
stack.push(new Pair(root.left, currNumber));
}
}
}
return rootToLeaf;
}
}