CODE 21: Path Sum

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本文介绍了一种方法，用于确定给定二叉树中是否存在从根节点到叶子节点的路径，使得路径上的节点值之和等于指定的总和。

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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.

	public boolean hasPathSum(TreeNode root, int sum) {
		// Start typing your Java solution below
		// DO NOT write main() function
		if (null == root) {
			return false;
		}
		return dfs(root, 0, sum);
	}

	private boolean dfs(TreeNode root, int current, int sum) {
		if (root.left == null && root.right == null) {
			if (current + root.val == sum) {
				return true;
			} else {
				return false;
			}
		}
		boolean is = false;
		if (root.left != null) {
			is = dfs(root.left, current + root.val, sum);
		}
		if (is) {
			return is;
		}
		if (root.right != null) {
			is = dfs(root.right, current + root.val, sum);
		}
		return is;
	}