Question

1022. Sum of Root To Leaf Binary Numbers

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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

# 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.
 * 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);
  }
}

/**
 * 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) {
    return dfs(root, 0);
  }

  int dfs(TreeNode* root, int t) {
    if (!root) return 0;
    t = (t << 1) | root->val;
    if (!root->left && !root->right) return t;
    return dfs(root->left, t) + dfs(root->right, t);
  }
};

/**
 * 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)
  }
}