LeetCode OJ 112. Path Sum

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.

代码如下：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if(root == null) return false;
        if(root.left==null && root.right==null){
            if(root.val==sum) return true;
            else return false;
        }
        
        boolean left = false;
        boolean right = false;
        if(root.left != null){
            left = hasPathSum(root.left, sum - root.val);
        }
        if(root.right != null){
            right = hasPathSum(root.right, sum - root.val);
        }
        
        if(left==true || right==true) return true;
        else return false;
    }
}