113. Path Sum II

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

Note: A leaf is a node with no children.

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]
]

难度：medium

题目：给定二叉树与一个数和sum，找出所有由根到叶结点的和为sum的路径

思路：递归（前序遍历）

Runtime: 2 ms, faster than 56.03% of Java online submissions for Path Sum II.
Memory Usage: 37.6 MB, less than 100.00% of Java online submissions for Path Sum II.

/**
 * 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>> result = new ArrayList<>();
        pathSum(root, sum, new Stack<>(), result);
        
        return result;
    }
    
    private void pathSum(TreeNode root, int sum, Stack<Integer> stack, List<List<Integer>> result) {
        if (null != root) {
            if (null == root.left && null == root.right) {
                if (sum == root.val) {
                    stack.push(root.val);
                    result.add(new ArrayList<>(stack));
                    stack.pop();
                }
            } else {
                stack.push(root.val);
                pathSum(root.left, sum - root.val, stack, result);
                pathSum(root.right, sum - root.val, stack, result);
                stack.pop();
            }
        }
    }
}