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剑指 Offer 34. 二叉树中和为某一值的路径 ​

题目 ​

给你二叉树的根节点 root 和一个整数目标和 targetSum ，找出所有 从根节点到叶子节点 路径总和等于给定目标和的路径。

叶子节点 是指没有子节点的节点。

示例 1：

输入：root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
输出：[[5,4,11,2],[5,8,4,5]]
输入：root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
输出：[[5,4,11,2],[5,8,4,5]]

示例 2：

输入：root = [1,2,3], targetSum = 5
输出：[]
输入：root = [1,2,3], targetSum = 5
输出：[]

示例 3：

输入：root = [1,2], targetSum = 0
输出：[]
输入：root = [1,2], targetSum = 0
输出：[]

提示：

  • 树中节点总数在范围 [0, 5000] 内
  • -1000 <= Node.val <= 1000
  • -1000 <= targetSum <= 1000

注意：本题与主站 113 题相同：https://leetcode-cn.com/problems/path-sum-ii/

题解 ​

java
public List<List<Integer>> pathSum(TreeNode root, int sum) {
    List<List<Integer>> result = new ArrayList<>();
    // 缓存累加值
    AtomicInteger count = new AtomicInteger();

    BiConsumer<TreeNode, Stack<Integer>> backtrack = new BiConsumer<TreeNode, Stack<Integer>>() {
        @Override
        public void accept(TreeNode node, Stack<Integer> stack) {
            if (node == null) {
                return;
            }
            // 栈存储每个节点值
            stack.push(node.val);
            count.addAndGet(node.val);
            // 满足条件的叶子节点
            if (node.left == null && node.right == null && count.get() == sum) {
                result.add(new ArrayList<>(stack));
            }

            this.accept(node.left, stack);
            this.accept(node.right, stack);
            // 回溯并计算累加值
            count.addAndGet(-stack.pop());
        }
    };

    backtrack.accept(root, new Stack<>());

    return result;
}
public List<List<Integer>> pathSum(TreeNode root, int sum) {
    List<List<Integer>> result = new ArrayList<>();
    // 缓存累加值
    AtomicInteger count = new AtomicInteger();

    BiConsumer<TreeNode, Stack<Integer>> backtrack = new BiConsumer<TreeNode, Stack<Integer>>() {
        @Override
        public void accept(TreeNode node, Stack<Integer> stack) {
            if (node == null) {
                return;
            }
            // 栈存储每个节点值
            stack.push(node.val);
            count.addAndGet(node.val);
            // 满足条件的叶子节点
            if (node.left == null && node.right == null && count.get() == sum) {
                result.add(new ArrayList<>(stack));
            }

            this.accept(node.left, stack);
            this.accept(node.right, stack);
            // 回溯并计算累加值
            count.addAndGet(-stack.pop());
        }
    };

    backtrack.accept(root, new Stack<>());

    return result;
}