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.

• 题目大意：从给定的二叉树中找到root - to -leaf的path，并且path上面的数字之和等于给定的sum。

public class PathSum {
    public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null)
            return false;

        //如果左右结点为null，此时到达叶子结点
        if (root.left == null && root.right == null && sum - root.val == 0)
            return true;

        //递归判断
        return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
    }

    public class TreeNode {
        int val;
        TreeNode left;
        TreeNode right;
        TreeNode(int x) { val = x; }
    }
}