110.Balanced Binary Tree (E)_110 isbalanced-优快云博客

本文介绍了一种判断二叉树是否平衡的算法，通过递归求各结点高度并检查左右子树高度差，确保树的高度平衡。适用于数据结构与算法学习。

Balanced Binary Tree (E)

Given a binary tree, determine if it is height-balanced.

For this problem, a height-balanced binary tree is defined as:

a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example 1:

Given the following tree [3,9,20,null,null,15,7]:

Return true.

Example 2:

Given the following tree [1,2,2,3,3,null,null,4,4]:

Return false.

题意

判断指定树是否为平衡二叉树，即对于树中任意一个结点，它的左子树和右子树高度之差不大于1。

思路

在递归求各结点高度的同时，判断当前结点的左子树和右子树高度之差是否大于1。

代码实现

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    boolean isBalanced = true;

    public boolean isBalanced(TreeNode root) {
        getHeight(root);
        return isBalanced;
    }

    private int getHeight(TreeNode x) {
        // 空结点时返回高度
        // 当已经确定该树不平衡时，无需再向下递归
        if (x == null || !isBalanced) {
            return 0;
        }

        int lHeight = getHeight(x.left);
        int rHeight = getHeight(x.right);

        if (Math.abs(lHeight - rHeight) > 1) {
            isBalanced = false;
        }

        return Math.max(lHeight, rHeight) + 1;
    }
}