110. 平衡二叉树

顶到底 效率太低了

class Solution {
  public boolean isBalanced(TreeNode root) {
      if(root == null)return true;
      if(Math.abs(func(root.left)-func(root.right))>1) return false;
      return isBalanced(root.left)&&isBalanced(root.right);
  }
  public int func(TreeNode node){
      if(node==null)return 0;
      else
          return Math.max(func(node.left)+1,func(node.right)+1);
  }
}

底到顶

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {
public boolean isBalanced(TreeNode root) {
  return recur(root) != -1;
}

private int recur(TreeNode root) {
  if(root == null) return 0;

  int left_value = recur(root.left);
  if(left_value == -1) return -1;

  int right_value = recur(root.right);
  if(right_value == -1) return -1;

  if(Math.abs(left_value-right_value)>1) return -1;
  return Math.max(left_value,right_value)+1;
}
}