LeetCode Oj 112. Path Sum 解題報告

LeetCode Oj 112. Path Sum 解題報告

112. Path Sum

My SubmissionsQuestionTotal Accepted:91133Total Submissions:295432Difficulty: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 andsum = 22,

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \      \
        7    2      1

return true, as there exist a root-to-leaf path5->4->11->2which sum is 22.

 

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顯然深搜求解，簡單高效。注意葉子節點的定義，當且僅當左右孩子都為空時，才是葉子節點。如果一個節點有一個孩子時，是不能作為leaf的。

我的AC代碼

 

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);
	}
}