Lintcode376-Binary Tree Path Sum-Easy

最新推荐文章于 2025-07-29 14:24:32 发布

转载最新推荐文章于 2025-07-29 14:24:32 发布 · 100 阅读

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原文链接：http://www.cnblogs.com/Jessiezyr/p/10656726.html

本文介绍了一种寻找二叉树中所有路径使得路径上节点值之和等于特定目标值的算法。通过递归方法实现，从根节点到叶子节点的所有合法路径将被收集并返回。

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376. Binary Tree Path Sum

Example

Example 1:

Input:
{1,2,4,2,3}
5
Output: [[1, 2, 2],[1, 4]]
Explanation:
The tree is look like this:
	     1
	    / \
	   2   4
	  / \
	 2   3
For sum = 5 , it is obviously 1 + 2 + 2 = 1 + 4 = 5

Example 2:

Input:
{1,2,4,2,3}
3
Output: []
Explanation:
The tree is look like this:
	     1
	    / \
	   2   4
	  / \
	 2   3
Notice we need to find all paths from root node to leaf nodes.
1 + 2 + 2 = 5, 1 + 2 + 3 = 6, 1 + 4 = 5 
There is no one satisfying it.

思路：

递归法，用helper方法，可以传更多参数

注意：

代码：

/*
     * @param root: the root of binary tree
     * @param target: An integer
     * @return: all valid paths
     */
    public List<List<Integer>> binaryTreePathSum(TreeNode root, int target) {
        List<List<Integer>> result = new ArrayList<>();
        ArrayList<Integer> path = new ArrayList<>();
        
        if (root == null) {
            return result;
        }
        path.add(root.val);
        helper(root, path, root.val, target, result);
        return result;
    }
    
    public void helper (TreeNode root,
                        ArrayList<Integer> path,
                        int sum,
                        int target,
                        List<List<Integer>> result) {
        //meet leaf
        if (root.right == null && root.left == null) {
            if (sum == target) {
                result.add(new ArrayList(path));
            }
            return;
        }
        // right node
        if (root.right != null) {
            path.add(root.right.val);
            helper(root.right, path, sum + root.right.val, target, result);
            path.remove(path.size() - 1);
        }
        //left node
        if (root.left != null) {
            path.add(root.left.val);
            helper(root.left, path, sum + root.left.val, target, result);
            path.remove(path.size() - 1);
        }
    }

转载于:https://www.cnblogs.com/Jessiezyr/p/10656726.html