方法1：BFS

时间复杂度：$O(n)$

空间复杂度：$O(n)$

解题思路

根节点入队
循环步骤1-3直到队列为空
1. 节点出队
2. 节点的左子节点入队
3. 节点的右子节点入队

双队列，分别存储节点和该节点的路径总和

https://leetcode.cn/problems/path-sum/solution/lu-jing-zong-he-by-leetcode-solution/

Java 代码

/**
 * 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 boolean hasPathSum(TreeNode root, int targetSum) {
        if (root == null) {
            return false;
        }
        Queue<TreeNode> q1 = new LinkedList<>();
        Queue<Integer> q2 = new LinkedList<>();
        q1.offer(root);
        q2.offer(root.val);
        while (!q1.isEmpty()) {
            TreeNode node = q1.poll();
            int tmp = q2.poll();
            if (node.left == null && node.right == null) {
                if (tmp == targetSum) return true;
                continue;
            }
            if (node.left != null) {
                q1.offer(node.left);
                q2.offer(tmp + node.left.val);
            }
            if (node.right != null) {
                q1.offer(node.right);
                q2.offer(tmp + node.right.val);
            }
        }
        return false;
    }
}