110. 平衡二叉树

110. 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 。

/**
 * 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 boolean isBalanced(TreeNode root) {
         if (root == null) {
            return true;
        } else {
            return Math.abs(height(root.left) - height(root.right)) <= 1 && isBalanced(root.left) && isBalanced(root.right);
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    &#125;

    public int height(TreeNode root) &#123;
        if (root == null) &#123;
            return 0;
        &#125; else &#123;
            return Math.max(height(root.left), height(root.right)) + 1;
        &#125;
    &#125;

// 作者：LeetCode-Solution
// 链接：https://leetcode-cn.com/problems/balanced-binary-tree/solution/ping-heng-er-cha-shu-by-leetcode-solution/

    &#125;
&#125;

/**
 * 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 boolean isBalanced(TreeNode root) {
        return height(root) >= 0;
    }

    public int height(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int leftHeight = height(root.left);
        int rightHeight = height(root.right);
        if (leftHeight == -1 || rightHeight == -1 || Math.abs(leftHeight - rightHeight) > 1) {
            return -1;
        } else {
            return Math.max(leftHeight, rightHeight) + 1;
        }
    }

// 作者：LeetCode-Solution
// 链接：https://leetcode-cn.com/problems/balanced-binary-tree/solution/ping-heng-er-cha-shu-by-leetcode-solution/

}