LeetCode | Path-Sum II

题目描述

 

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 andsum = 22,

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \    / \
        7    2  5   1

return

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

题目意思: 求根节点到叶子几点路径和为sum的全部路径。

import java.util.*;
/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    ArrayList<ArrayList<Integer>> listPath = new ArrayList<>();
    public ArrayList<ArrayList<Integer>> pathSum(TreeNode root, int sum) {
        
        getPath(root, sum, new ArrayList<Integer>());
        return listPath;
    }
    
    public void getPath(TreeNode root, int sum, ArrayList<Integer> path) {
        if (root == null) {
            return;
        }
        sum -= root.val; //首先将sum减去当前节点的值
        if (root.left == null && root.right==null) {//判断该节点是否为叶子节点
            if (sum == 0) {//该节点为叶子且sum为0,满足条件
                path.add(root.val);
                listPath.add(new ArrayList<Integer>(path));
                path.remove(path.size()-1);
            }
            return;
        }
        path.add(root.val); //path添加当前值
        getPath(root.left, sum, path);    //左子树进行上面操作
        getPath(root.right, sum, path); //右子树进行上面操作
        path.remove(path.size()-1);    //该节点的左右子树都操作完后，删除该节点在path上的值
    }
}