package leetcode_1To300;

import java.util.Stack;

/**
 * 本代码来自 Cspiration，由 @Cspiration 提供
 * 题目来源：http://leetcode.com
 * - Cspiration 致力于在 CS 领域内帮助中国人找到工作，让更多海外国人受益
 * - 现有课程：Leetcode Java 版本视频讲解（1-900题）（上）（中）（下）三部
 * - 算法基础知识（上）（下）两部；题型技巧讲解（上）（下）两部
 * - 节省刷题时间，效率提高2-3倍，初学者轻松一天10题，入门者轻松一天20题
 * - 讲师：Edward Shi
 * - 官方网站：https://cspiration.com
 * - 版权所有，转发请注明出处
 */
public class _112_PathSum {
    /**
     * 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.

     time : O(n);
     space : O(n);
     * @param root
     * @param sum
     * @return
     */
    public static boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) return false;
        if (root.left == null && root.right == null) {
            return sum == root.val;
        }
        return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
    }

    public static boolean hasPathSum2(TreeNode root, int sum) {
        if (root == null) return false;
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        while (!stack.isEmpty()) {
            TreeNode cur = stack.pop();
            if (cur.left == null && cur.right == null) {
                if (cur.val == sum) {
                    return true;
                }
            }
            if (cur.right != null) {
                stack.push(cur.right);
                cur.right.val += cur.val;
            }
            if (cur.left != null) {
                stack.push(cur.left);
                cur.left.val += cur.val;
            }
        }
        return false;
    }
}