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.

SOLUTION 1:

相当简单的题目，用递归解决。base case: 如果root是叶子，并且sum为root的值，返回true.

否则在左右分别 找一下就好（调用递归），任何一边返回true都可以的。

1 public boolean hasPathSum(TreeNode root, int sum) { 2         if (root == null) { 3             return false; 4         } 5  6         if (root.left == null && root.right == null && sum == root.val) { 7             return true; 8         } 9 10         return hasPathSum(root.left, sum - root.val)11                 || hasPathSum(root.right, sum - root.val);12     }