leetcode 113. Path Sum II

最新推荐文章于 2023-11-14 10:36:43 发布

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本文介绍了一种算法，用于在给定的二叉树中找到所有从根节点到叶子节点的路径，使得这些路径上的元素之和等于指定的数值。通过实例演示了如何实现这一算法。

Question

Given a binary tree and a sum, find all root-to-leaf paths where each path’s sum equals the given sum.

For example:
Given the below binary tree and sum = 22,

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \    / \
        7    2  5   1

return
[
[5,4,11,2],
[5,8,4,5]
]

Code

public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> rst = new ArrayList<>();
        List<Integer> solution = new ArrayList<Integer>();

        findSum(rst, solution, root, sum);
        return rst;
    }

    private void findSum(List<List<Integer>> result, List<Integer> solution, TreeNode root, int sum) {
        if (root == null) {
            return;
        }

        sum -= root.val;

        if (root.left == null && root.right == null) {
            if (sum == 0) {
                solution.add(root.val);
                result.add(new ArrayList<Integer>(solution));
                solution.remove(solution.size() - 1);
            }
            return;
        }

        solution.add(root.val);
        findSum(result, solution, root.left, sum);
        findSum(result, solution, root.right, sum);
        solution.remove(solution.size() - 1);
    }