112. Path Sum
說明：
給予一個二元樹和一個數字target，從root到葉節點的路徑稱為path，若存在任意path中
所有數字和等於target返回true，否則返回false。
Example 1:
https://assets.leetcode.com/uploads/2021/01/18/pathsum1.jpg
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.
解法：
1.DFS中序遍歷並更新targetSum的值
2.若root是葉子節點且等於targetSum返回true
Java Code：
class Solution {
public boolean hasPathSum(TreeNode root, int targetSum) {
if(root == null) return false;
if(targetSum == root.val && root.left == null && root.right == null)
return true;
return hasPathSum(root.left, targetSum - root.val) ||
hasPathSum(root.right, targetSum - root.val);
}
}
樹 = 溫柔善良