112 Path Sum

Given the root of a binary tree and an integer targetSum, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum.

A leaf is a node with no children.

Example 1:

Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
Output: true
Explanation: The root-to-leaf path with the target sum is shown.

Example 2:

Input: root = [1,2,3], targetSum = 5
Output: false
Explanation: There two root-to-leaf paths in the tree:
(1 --> 2): The sum is 3.
(1 --> 3): The sum is 4.
There is no root-to-leaf path with sum = 5.

Example 3:

Input: root = [], targetSum = 0
Output: false
Explanation: Since the tree is empty, there are no root-to-leaf paths.

Solution:

题目要求的能拐弯吗? 不能-> case b

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

       if (root.left == null && root.right == null && root.val == targetSum){
           return true;
       }

       int sum = 0;
       return helper(root, sum, targetSum);
    }

    public static boolean helper(TreeNode root, int sum, int targetSum){
        // base case
        if (root == null){
            return false;
        }

        sum = sum + root.val;

        if (root.left == null && root.right == null){
            if (sum == targetSum){
                return true;
            }else{
                return false;
            }
        }

        if (root.left == null){
            return helper(root.right, sum, targetSum);
        }

        if (root.right == null){
            return helper(root.left, sum, targetSum);
        }

        return (helper(root.left, sum, targetSum) || helper(root.right, sum, targetSum));
    }
}

// TC: O(n)
// SC: O(n)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean hasPathSum(TreeNode root, int targetSum) {
        // base case 
        if (root == null){
            return false;
        }

        targetSum = targetSum - root.val;
        if (targetSum == 0 && root.left == null && root.right == null){
            return true;
        }

        boolean left = hasPathSum(root.left, targetSum);
        boolean right = hasPathSum(root.right, targetSum);

        if (left == true || right == true){
            return true;
        }else{
            return false;
        }
    }
}