题解 | #二叉树中和为某一值的路径(一)#

递归

/*
 * public class TreeNode {
 *   int val = 0;
 *   TreeNode left = null;
 *   TreeNode right = null;
 * }
 */

public class Solution {
    /**
     * 
     * @param root TreeNode类 
     * @param sum int整型 
     * @return bool布尔型
     */
    public boolean hasPathSum (TreeNode root, int sum) {
        // write code here
        if(root==null) return false;
        sum=sum-root.val;
        if(sum==0&&root.right==null&&root.left==null) return true;
        return hasPathSum(root.left,sum)||hasPathSum(root.right,sum);
    }
}