113. Path Sum II

题目分析

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 and sum = 22,

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

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

代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> paths = new ArrayList<>();
        ArrayList<Integer> path = new ArrayList<>();
        if(root == null) {
            return paths;
        }
        helper(root, sum, path, paths);
        return paths;
    }
    public void helper(TreeNode root, int sum, List<Integer> path, List<List<Integer>> paths) {
        // 判断是不是叶子结点
        if(root == null) {
            return;
        }
        sum -= root.val;
        path.add(root.val);
        if(root.left == null && root.right == null) {
            if(0 == sum) { 
                paths.add(new ArrayList<Integer>(path));
            }
        } else {
            if(root.left != null) helper(root.left, sum, path, paths);
            if(root.right != null) helper(root.right, sum, path, paths);
        }
        path.remove(path.size() - 1);
    }
}