leetcode 112. 路径总和

package com.example.lettcode.dailyexercises;

/**
 * @Class HasPathSum
 * @Description 112. 路径总和
 * 给定一个二叉树和一个目标和，判断该树中是否存在根节点到叶子节点的路径，
 * 这条路径上所有节点值相加等于目标和。
 * <p>
 * 说明: 叶子节点是指没有子节点的节点。
 * 示例:
 * 给定如下二叉树，以及目标和 sum = 22，
 * <p>
 * 5
 * / 
 * 4   8
 * /   / 
 * 11  13  4
 * /        
 * 7    2      1
 * 返回 true, 因为存在目标和为 22 的根节点到叶子节点的路径 5->4->11->2。
 * @Author
 * @Date 2020/7/7
 **/
public class HasPathSum {
    static class TreeNode {
        int val;
        TreeNode left;
        TreeNode right;

        TreeNode(int x) {
            val = x;
        }
    }
}

/**
 * 递归方式
 */
public static boolean hasPathSum(TreeNode root, int sum) {
	if (root == null) {
		return false;
	}

	// 如果是叶子节点，判断节点的值与sum是否相等
	if (root.left == null && root.right == null) return root.val == sum;

	return hasPathSum(root.left, sum - root.val)
			|| hasPathSum(root.right, sum - root.val);
}

// 测试用例
    public static void main(String[] args) {
        TreeNode root = new TreeNode(5);
        TreeNode treeNode1 = new TreeNode(4);
        TreeNode treeNode2 = new TreeNode(8);
        root.left = treeNode1;
        root.right = treeNode2;
        TreeNode treeNode3 = new TreeNode(11);
        treeNode1.left = treeNode3;
        TreeNode treeNode4 = new TreeNode(13);
        TreeNode treeNode5 = new TreeNode(4);
        treeNode2.left = treeNode4;
        treeNode2.right = treeNode5;
        TreeNode treeNode6 = new TreeNode(7);
        TreeNode treeNode7 = new TreeNode(2);
        treeNode3.left = treeNode6;
        treeNode3.right = treeNode7;
        TreeNode treeNode8 = new TreeNode(1);
        treeNode5.right = treeNode8;
        int sum = 22;
        boolean ans = hasPathSum(root, sum);
        System.out.println("HasPathSum demo01 result:" + ans);
    }