Path Sum
<p>Given the <code>root</code> of a binary tree and an integer <code>targetSum</code>, return <code>true</code> if the tree has a <strong>root-to-leaf</strong> path such that adding up all the values along the path equals <code>targetSum</code>.</p>

<p>A <strong>leaf</strong> is a node with no children.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0112.Path%20Sum/images/pathsum1.jpg" style="width: 500px; height: 356px;" />
<pre>
<strong>Input:</strong> root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
<strong>Output:</strong> true
<strong>Explanation:</strong> The root-to-leaf path with the target sum is shown.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0112.Path%20Sum/images/pathsum2.jpg" />
<pre>
<strong>Input:</strong> root = [1,2,3], targetSum = 5
<strong>Output:</strong> false
<strong>Explanation:</strong> There are two root-to-leaf paths in the tree:
(1 --&gt; 2): The sum is 3.
(1 --&gt; 3): The sum is 4.
There is no root-to-leaf path with sum = 5.
</pre>

<p><strong class="example">Example 3:</strong></p>

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

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[0, 5000]</code>.</li>
	<li><code>-1000 &lt;= Node.val &lt;= 1000</code></li>
	<li><code>-1000 &lt;= targetSum &lt;= 1000</code></li>
</ul>
Tree; Depth-First Search; Breadth-First Search; Binary Tree
// 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 has_path_sum(root: Option<Rc<RefCell<TreeNode>>>, target_sum: i32) -> bool {
        match root {
            None => false,
            Some(node) => {
                let mut node = node.borrow_mut();
                // 确定叶结点身份
                if node.left.is_none() && node.right.is_none() {
                    return target_sum - node.val == 0;
                }
                let val = node.val;
                Self::has_path_sum(node.left.take(), target_sum - val)
                    || Self::has_path_sum(node.right.take(), target_sum - val)
            }
        }
    }
}
