题解 | #判断是不是平衡二叉树#
判断是不是平衡二叉树
https://www.nowcoder.com/practice/8b3b95850edb4115918ecebdf1b4d222
import java.util.*; /* * public class TreeNode { * int val = 0; * TreeNode left = null; * TreeNode right = null; * public TreeNode(int val) { * this.val = val; * } * } */ public class Solution { /** * 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 * * * @param pRoot TreeNode类 * @return bool布尔型 */ public boolean IsBalanced_Solution (TreeNode pRoot) { // write code here return getHeight(pRoot) != -1; } private int getHeight(TreeNode node) { if (node == null)return 0; // 对当前节点,递归判断其左右子树是否为平衡二叉树 int leftCnt = getHeight(node.left); if(leftCnt == -1)return -1; int rightCnt = getHeight(node.right); if(rightCnt == -1)return -1; // 判断左右子树高度差是否小于等于1,是则返回当前节点的高度 return Math.abs(leftCnt - rightCnt) > 1 ? -1 : Math.max(leftCnt, rightCnt)+1; } }
递归判断,先判断左右子树是否为平衡二叉树