376. Binary Tree Path Sum【LintCode java】

Description

Given a binary tree, find all paths that sum of the nodes in the path equals to a given number target.

A valid path is from root node to any of the leaf nodes.

Example

Given a binary tree, and target = 5:

     1
    / \
   2   4
  / \
 2   3

return

[
  [1, 2, 2],
  [1, 4]
]

解题：给一个二叉树，找到和是给定目标值的支路。如果不用栈，需要考虑一下回溯算法的运用，代码如下：

/**
 * 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
     */
    int target = 0;
    List<List<Integer>> res = new ArrayList<>();
    public List<List<Integer>> binaryTreePathSum(TreeNode root, int target) {
        // write your code here
        if(root == null){
            return res;
        }
        this.target = target;
        List<Integer>path = new ArrayList<>();
        helper(path, root, 0);
        return res;
    }
    public void helper(List<Integer>path, TreeNode root, int sum){
        sum +=root.val;
        path.add(root.val);
        //如果到叶子结点了，并且和为target
        if(root.left == null && root.right == null && sum == target){
            ArrayList<Integer>temp = new ArrayList<>();
            temp.addAll(path);//复制一份
            res.add(temp);
        }
        if(root.left != null){
            helper(path, root.left, sum);
        }
        if(root.right != null){
            helper(path, root.right, sum);
        }
        path.remove(path.size() - 1);
    }
}