Viiisit [LeetCode] - 112. Path Sum

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Problem - Path Sum

Given the root of a binary tree and an integer targetSum, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum.

A leaf is a node with no children.

翻譯蒟蒻

給定一個二元樹的根節點和一個整數目標總和（targetSum），如果該樹具有從根到葉節點的路徑，使得沿著該路徑的所有值相加等於目標總和，則返回true；否則返回false。

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• Example 1:

  Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
  Output: true
  Explanation: The root-to-leaf path with the target sum is shown.

• Example 2:

  Input: root = [1,2,3], targetSum = 5
  Output: false
  Explanation: There two root-to-leaf paths in the tree:
  (1 --> 2): The sum is 3.
  (1 --> 3): The sum is 4.
  There is no root-to-leaf path with sum = 5.

• Example 3:

  Input: root = [], targetSum = 0
  Output: false
  Explanation: Since the tree is empty, there are no root-to-leaf paths.

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Solution - JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} targetSum
 * @return {boolean}
 */
var hasPathSum = function(root, targetSum) {
      if (root === null) {
          return false;
      }
  
      if (root.left === null && root.right === null) {
          // 檢查目標和減去葉節點的值是否為 0，如果為 0，則找到了一條路徑，其節點值的和等於目標和，返回 true
          return targetSum - root.val === 0;
      }

      return hasPathSum(root.left, targetSum - root.val) || hasPathSum(root.right, targetSum - root.val);
};

Solution - Ruby

def has_path_sum(root, target_sum)
    return false if root.nil?
    return true if root.left.nil? && root.right.nil? && root.val == target_sum
    return has_path_sum(root.left, target_sum - root.val) || has_path_sum(root.right, target_sum - root.val)
end

Solution - PHP

function hasPathSum($root, $targetSum) {
    if ($root == null) return false;
    if ($root->left == null && $root->right == null) return $targetSum == $root->val;
    return $this->hasPathSum($root->left, $targetSum - $root->val) || $this->hasPathSum($root->right, $targetSum - $root->val);
}

LeetCode 傳送門 - Path Sum