LeetCode Oj 112. Path Sum 解题报告

112. Path Sum

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Question

Total Accepted: 91133 Total
Submissions: 295432 Difficulty: 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.

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显然深搜求解，简单高效。注意叶子节点的定义，当且仅当左右孩子都为空时，才是叶子节点。如果一个节点有一个孩子时，是不能作为leaf的。

我的AC代码

public class PathSum {
	static Boolean ok = false;
	/**
	 * @param args
	 */
	public static void main(String[] args) {
		TreeNode treeNode = new TreeNode(1);
		TreeNode t1 = new TreeNode(2);
		System.out.println(hasPathSum(treeNode, 1));
		System.out.println(hasPathSum(null, 0));
		treeNode.left = t1;
		System.out.println(hasPathSum(treeNode, 1));

	}

	public static boolean hasPathSum(TreeNode root, int sum) {
		if(root == null) return false;
		ok = false;
        dfs(root, sum, 0);
        return ok;
    }

	public static void dfs(TreeNode root, int sum, int s) {
		if(ok == true) return;
		if(root.left == null && root.right == null) {
			if (s + root.val == sum) {
				ok = true;
			}
			return;
		}

		if(root.left != null) dfs(root.left, sum, s + root.val);
		if(root.right != null) dfs(root.right, sum, s + root.val);
	}
}