leetcode-java-110. Balanced Binary Tree

/*平衡二叉树：它是一 棵空树或它的左右两个子树的高度差的绝对值不超过11.递归判断二叉树是否是平衡二叉树2.递归找二叉树高度,求高度也是用递归    递归停止条件：    递归到叶子节点时终止，即root.left==null && root.right == null时,返回1；root==null时，返回0。 */ public class Solution {     public boolean isBalanced(TreeNode root) {         if(root == null) {             return true;         }         int l = height(root.left);         int r = height(root.right);         //若本节点满足平衡二叉树，判断左右子树是否满足         if(l-r >= -1 && l-r <= 1) {             return isBalanced(root.left) && isBalanced(root.right);         } else {             return false;         }     }     public int height(TreeNode root) {         int leftHeight,             rightHeight;         if(root == null) {             return 0;         }         if(root.left == null && root.right == null) {             return 1;         }         leftHeight = height(root.left);         rightHeight = height(root.right);         return leftHeight>=rightHeight ? leftHeight+1:rightHeight+1;     } }