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1022. Sum of Root To Leaf Binary Numbers

Description

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.

Solutions

Solution 1: Recursion

Thinking

Listing every root-to-leaf path and converting the bits is fine for \(n\le 1000\), but the paths need extra storage. Walking downward, the path value updates as \(t\leftarrow 2t+\textit{val}\) and is complete at a leaf.

A null node contributes \(0\). A node with no children is a leaf and returns the current \(t\); otherwise both subtrees receive the same \(t\) and we add the results.

DFS carries the path value and visits each node once.

We design a recursive function \(\text{dfs}(root, t)\), which takes two parameters: the current node \(root\) and the binary number \(t\) corresponding to the parent node of the current node. The return value of the function is the sum of binary numbers represented by paths from the current node to leaf nodes. The answer is \(\textrm{dfs}(root, 0)\).

The logic of the recursive function is as follows:

  • If the current node \(root\) is null, return \(0\); otherwise, calculate the binary number \(t\) corresponding to the current node, i.e., \(t = t \ll 1 | root.val\).
  • If the current node is a leaf node, return \(t\); otherwise, return the sum of \(\textrm{dfs}(root.left, t)\) and \(\textrm{dfs}(root.right, t)\).

The time complexity is \(O(n)\), and the space complexity is \(O(n)\), where \(n\) is the number of nodes in the binary tree. Each node is visited once; the recursion stack requires \(O(n)\) space.

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# 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: Optional[TreeNode]) -> int:
        def dfs(root: Optional[TreeNode], x: int) -> int:
            if root is None:
                return 0
            x = x << 1 | root.val
            if root.left == root.right:
                return x
            return dfs(root.left, x) + dfs(root.right, x)

        return dfs(root, 0)

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/**
 * 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 x) {
        if (root == null) {
            return 0;
        }
        x = x << 1 | root.val;
        if (root.left == root.right) {
            return x;
        }
        return dfs(root.left, x) + dfs(root.right, x);
    }
}

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int sumRootToLeaf(TreeNode* root) {
        auto dfs = [&](this auto&& dfs, TreeNode* root, int x) -> int {
            if (!root) {
                return 0;
            }
            x = x << 1 | root->val;
            if (root->left == root->right) {
                return x;
            }
            return dfs(root->left, x) + dfs(root->right, x);
        };
        return dfs(root, 0);
    }
};

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

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/**
 * 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 = (node: TreeNode | null, x: number): number => {
        if (node === null) {
            return 0;
        }

        x = (x << 1) | node.val;

        if (node.left === null && node.right === null) {
            return x;
        }

        return dfs(node.left, x) + dfs(node.right, x);
    };

    return dfs(root, 0);
}

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// 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::cell::RefCell;
use std::rc::Rc;

impl Solution {
    pub fn sum_root_to_leaf(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
        fn dfs(node: Option<Rc<RefCell<TreeNode>>>, x: i32) -> i32 {
            if let Some(n) = node {
                let n_ref = n.borrow();
                let x = (x << 1) | n_ref.val;

                if n_ref.left.is_none() && n_ref.right.is_none() {
                    return x;
                }

                dfs(n_ref.left.clone(), x) + dfs(n_ref.right.clone(), x)
            } else {
                0
            }
        }

        dfs(root, 0)
    }
}

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