Leetcode: Path Sum II

寻找二叉树路径之和

最新推荐文章于 2024-09-06 16:32:48 发布

原创最新推荐文章于 2024-09-06 16:32:48 发布 · 583 阅读

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本文介绍了一种算法，用于查找二叉树中所有从根节点到叶子节点的路径，使得路径上节点值的总和等于给定的目标值。通过递归方式实现，包括路径的构建与回溯。

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]
]

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<ArrayList<Integer>> pathSum(TreeNode root, int sum) {
		// Start typing your Java solution below
		// DO NOT write main() function
		ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
		ArrayList<Integer> path = new ArrayList<Integer>();
		traversal(res, path, root, sum);
		return res;
	}
	
	public void traversal(ArrayList<ArrayList<Integer>> res, ArrayList<Integer> path, TreeNode root, int sum){
		if(root == null)
			return;
		path.add(root.val);
		sum = sum - root.val;
		if(root.left == null && root.right == null && sum == 0)
			res.add(new ArrayList<Integer>(path));
		traversal(res, path, root.left, sum);
		traversal(res, path, root.right, sum);
		path.remove(path.size() - 1);
	}
}

555...代码一遍过，不容易啊

发现其实还可以剪枝，如果过程中sum已经小于零了，直接这条路径不再继续return