面试题34. 二叉树中和为某一值的路径(dfs)

输入一棵二叉树和一个整数,打印出二叉树中节点值的和为输入整数的所有路径。从树的根节点开始往下一直到叶节点所经过的节点形成一条路径。

示例:
给定如下二叉树,以及目标和 sum = 22,

5
/
4 8
/ /
11 13 4
/ /
7 2 5 1
返回:

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

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    List<List<Integer>> ans = new ArrayList<>();
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        dfs(root, sum, 0, new ArrayList<>());
        return ans;
    }

    public void dfs(TreeNode root, int sum, int value, List<Integer> list)
    {
        if (root == null) return;
        list.add(root.val);
        if (sum == value + root.val && root.left == null && root.right == null)
        {
            ans.add(new ArrayList<>(list));
        }
        dfs (root.left, sum, value + root.val, list);
        dfs (root.right, sum, value + root.val,list);
        list.remove(list.size() - 1);
    }   
}
原文地址:https://www.cnblogs.com/ziytong/p/13163318.html