Path Sum leetcode

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

For example:
Given the below binary tree and sum = 22,

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

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

递归(分治)方法来做，递归条件是看左子树或者右子树有没有满足条件的路径，也就是子树路径和等于当前sum减去当前节点的值。结束条件是如果当前节点是空的，则返回false，如果是叶子，那么如果剩余的sum等于当前叶子的值，则找到满足条件的路径，返回true。算法的复杂度是输的遍历，时间复杂度是O(n)，空间复杂度是O(logn)。

public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) {
            return false;
        }
        if (root.left == null && root.right == null && root.val == sum) {
            return true;
        }
        return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
    }
}