给定一个二叉树,判断它是否是高度平衡的二叉树。
本题中,一棵高度平衡二叉树定义为:
一个二叉树每个节点 的左右两个子树的高度差的绝对值不超过 1 。
示例 1:
输入:root = [3,9,20,null,null,15,7]
输出:true
示例 2:
输入:root = [1,2,2,3,3,null,null,4,4]
输出:false
示例 3:
输入:root = []
输出:true
class Solution {
public int height(TreeNode root){
if(root == null) return 0;
int leftheight = height(root.left);
int rightheight = height(root.right);
return (leftheight > rightheight) ? (leftheight+1):(rightheight+1);
}
public boolean isBalanced(TreeNode root) {
if(root == null) return true;
int left = height(root.left);
int right = height(root.right);
return Math.abs(left - right) <= 1 && isBalanced(root.left) && isBalanced(root.right);
}
}
该代码的时间复杂度为O(n*n),将代码可以优化如下:
class Solution {
public int height(TreeNode root){
if(root == null) {return 0;}
int leftheight = height(root.left);
int rightheight = height(root.right);
if(leftheight >= 0 && rightheight >= 0 && Math.abs(leftheight - rightheight) <= 1){
return Math.max(leftheight,rightheight) + 1;
}else{
//说明树已经不平衡了
return -1;
}
}
public boolean isBalanced(TreeNode root) {
if(root == null) {return true;}
return height(root) >= 0;
}
}