package com.leecode.tree;

public class code00hasPathSum {

code00hasPathSum left;

code00hasPathSum right;

int val;

//取值函数

code00hasPathSum(int val) {

this.val = val;

}

public static void main(String[] args) {

code00hasPathSum root = new code00hasPathSum(5);

code00hasPathSum left1 = new code00hasPathSum(4);

code00hasPathSum left2 = new code00hasPathSum(11);

code00hasPathSum left3 = new code00hasPathSum(7);

code00hasPathSum left4 = new code00hasPathSum(2);

code00hasPathSum right1 = new code00hasPathSum(8);

code00hasPathSum right2 = new code00hasPathSum(13);

code00hasPathSum right3 = new code00hasPathSum(4);

code00hasPathSum right4 = new code00hasPathSum(1);

// 构建这棵树

root.left = left1;

left1.left = left2;

left2.left = left3;

left2.right = left4;

root.right = right1;

right1.left = right2;

right1.right = right3;

right3.right = right4;

// 传入根节点

int sum = 22;

boolean res = hasPathSum(root, sum);

if (res) {

System.out.printf("这棵树有和为%d的路径", sum);

} else {

System.out.printf("这棵树没有和为%d的路径", sum);

}

}

// 路径之和方法

public static boolean hasPathSum(code00hasPathSum root, int sum) {

if (root == null) {

return false;

}

sum -= root.val;

if (root.left == null && root.right == null) {

return (sum == 0);

}

// 开始递归

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

}

}