112. Path Sum

Easy

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.

自己写的比较繁琐复杂, 思路是记录每一条从根节点到叶节点的路径，直到到达叶节点，才计算整条路径的root to leaf的sum. 如果有sum == target，则直接返回true. 如果没有遇到，则继续用DFS搜索左右子树，期间用到backtracking。左右子树只要有一条路径满足条件就返回true.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        List path = new ArrayList<>();
        return helper(root, sum, path);
    }
    
    private boolean helper(TreeNode root, int target, List path){
        if (root == null){
            return false;
        }
        path.add(root.val);
        int sum = 0;
        if (root.left == null && root.right == null){
            for (int i = 0; i < path.size(); i++){
                sum += path.get(i);   
            }
            if (sum == target){
                return true;
            }
        }
        path.remove(path.size() - 1);
        return helper(root.left, target, path) || helper(root.right, target, path);
    }
}

精简之后可以写成：

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