- 平衡二叉树
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
提示：
树中的节点数在范围 [0, 5000] 内
-104 <= Node.val <= 104
方法一：时间复杂度为0（n²）
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public int maxDepth(TreeNode root) {
if(root == null) return 0;
int leftHeight = maxDepth(root.left);
int rightHeight = maxDepth(root.right);
return leftHeight > rightHeight ?
leftHeight +1 : rightHeight +1;
}
public boolean isBalanced(TreeNode root) {
if(root == null) {
return true;
}
int leftHeight = maxDepth(root.left);
int rightHeight = maxDepth(root.right);
return Math.abs(leftHeight-rightHeight) <= 1
&& isBalanced(root.left) && isBalanced(root.right);
}
}
方法二：时间复杂度为0（n）
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public int hight(TreeNode root) {
if(root == null) return 0;
int leftHight = hight(root.left);
int rightHight = hight(root.right);
// if(leftHight >= 0 && rightHight >= 0 &&
// Math.abs(leftHight-rightHight) <= 1) {
// return Math.max(leftHight,rightHight) + 1;
// }else {
// return -1;
// }
if(leftHight == -1 || rightHight == -1 ||
Math.abs(leftHight-rightHight) > 1) {
return -1;
}else {
return Math.max(leftHight,rightHight) + 1;
}
}
public boolean isBalanced(TreeNode root) {
if(hight(root) >= 0) {
return true;
}
return false;
//return hight(root) >= 0;
}
}