描述

给定一个二叉树，找出所有路径中各节点相加总和等于给定 目标值 的路径。
一个有效的路径，指的是从根节点到叶节点的路径。

样例

给定一个二叉树，和 目标值 = 5:
     1
    / \
   2   4
  / \
 2   3
返回：
[[1, 2, 2],
  [1, 4]]

代码实现

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root the root of binary tree
     * @param target an integer
     * @return all valid paths
     */
   // List<List<Integer>> result = new ArrayList<>();
    public List<List<Integer>> binaryTreePathSum(TreeNode root, int target) {
        List<List<Integer>> result = new ArrayList<>();
        if (root == null) {
            return result;
        }
        List<Integer> path = new ArrayList<>();
        path.add(root.val);
        dfs(root, root.val, target, path, result);
        return result;
    }
    private void dfs(TreeNode root, int sum, int target, List<Integer> path,
                    List<List<Integer>> result) {
        // meat leaf
        if (root.left == null && root.right == null) {
            if (sum == target) {
                result.add(new ArrayList<>(path));
            }
            return;
        }
        // go left
        if (root.left != null) {
            path.add(root.left.val);
            // not dfs(root, sum+root.left.val, target, path, result);
            dfs(root.left, sum+root.left.val, target, path, result);
           //去掉上一次加入的节点 dfs
           path.remove(path.size()-1);
        }
        //go right 
        if (root.right != null) {
            path.add(root.right.val);
            //not dfs(root, sum+root.right.val, target, path, result);
            dfs(root.right, sum+root.right.val, target, path, result);
            path.remove(path.size()-1);
        }
    }
}