二叉树的路径和

题目描述:  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。

思路: 二叉树前序遍历.
1. 若当前节点为空,则直接返回
2. 当前节点不为空,从根节点的路径和加上该节点的值得到根节点到该节点的和
3. 当前节点的左儿子和右儿子为空,判断当前和是否与目标和一致,若一致,则保存路径,然后返回
4. 当前节点存在左儿子或者右儿子,将当前节点加入到路径中,然后分别向左子树和右子树计算,左右都计算完成后,将该节点从路径中除去.

import java.util.ArrayList;import java.util.List;import java.util.Scanner;/** * Created by parting_soul on 17-6-28. * 二叉树路径和 * <p> * 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。 */public class BinaryTreeAnd {    public static void main(String[] args) {        Scanner in = new Scanner(System.in);        TreeNode root = buildTreeByPreOrder(in);        preOrder(root);        System.out.println("输入目标值:");        int target = in.nextInt();        in.close();        Solution solution = new Solution();        solution.binaryTreePathSum(root, target);        System.out.println(solution.lists);    }    /**     * 输入前序遍历结果建立二叉树,每个字符间有空格,空节点以-1表示     *     * @param in     */    private static TreeNode buildTreeByPreOrder(Scanner in) {        int value = in.nextInt();        TreeNode node = null;        if (value != -1) {            node = new TreeNode(value);            node.left = buildTreeByPreOrder(in);            node.right = buildTreeByPreOrder(in);        }        return node;    }    /**     * 前序遍历     *     * @param node     */    private static void preOrder(TreeNode node) {        if (node == null) return;        System.out.print(node.val);        preOrder(node.left);        preOrder(node.right);    }    static class Solution {        ArrayList<List<Integer>> lists = new ArrayList<>();        /**         * @param root   the root of binary tree         * @param target an integer         * @return all valid paths         */        public List<List<Integer>> binaryTreePathSum(TreeNode root, int target) {            ArrayList<Integer> path = new ArrayList<>();            preOrderSum(root, path, lists, target, 0);            return lists;        }        private void preOrderSum(TreeNode node, ArrayList<Integer> path, ArrayList<List<Integer>> lists, int target, int sum) {            if (node == null) {                return;            }//            System.out.println(path);            sum += node.val;//            //当前路径和比目标值大,不往下走//            if (sum > target) return;            if (node.left == null && node.right == null) {                if (sum == target) {                    //lists.add(path);                    ArrayList<Integer> pathI = new ArrayList<>();                    for (int i = 0; i < path.size(); i++) {                        int temp = path.get(i);                        pathI.add(temp);                    }                    lists.add(pathI);                    pathI.add(node.val);                }                return;            }            path.add(node.val);            int len = path.size();            preOrderSum(node.left, path, lists, target, sum);            preOrderSum(node.right, path, lists, target, sum);            path.remove(len - 1);        }    }    static class TreeNode {        public int val;        public TreeNode left, right;        public TreeNode(int val) {            this.val = val;            this.left = this.right = null;        }    }}