二叉树的平衡判断（Java）

最新推荐文章于 2025-06-06 15:01:27 发布

Lanzijingshizi

最新推荐文章于 2025-06-06 15:01:27 发布

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分类专栏：【面试笔试】-找实习找工作文章标签： Java

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本文介绍了一种用于检查二叉树是否为平衡二叉树的算法。平衡二叉树的定义是任意节点的左右子树深度相差不超过1。通过递归计算每个节点的左右子树深度，并比较深度差来实现检查。

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问题描述：
如果某二叉树任意节点的左右子树的深度相差不能超过1，那么它就是一颗平衡二叉树。请事先未完成的checkBalance函数，判断给定的输入节点是否为平衡二叉树。
bool checkBalance(Node* root)

class TreeNode{
	int val;
	TreeNode left;
	TreeNode right;
	TreeNode(int val){
		this.val = val;
		left = null;
		right = null;
	}
	TreeNode(int val,TreeNode left,TreeNode right){
		this.val = val;
		this.left = left;
		this.right = right;
	}
}

public class Solution1 {
	
	boolean checkBalance(TreeNode root) {
		if(root == null)
			return true;
		
		int right = deep(root.right);
		int left = deep(root.left);
		if(right-left > 1 || left - right > 1)
			return false;
		
		return checkBalance(root.right) && checkBalance(root.left);
	}
	
	int deep(TreeNode root) {
		if(root == null)
			return 0;
		int deepRight = deep(root.right);
		int deepLeft = deep(root.left);
		return 1 + (deepRight>deepLeft?deepRight:deepLeft);
	}

	public static void main(String[] args) {
		
		TreeNode val_2 = new TreeNode(2);
		TreeNode val_4 = new TreeNode(4);
		TreeNode val_3 = new TreeNode(3,val_2,val_4);
		//TreeNode root = new TreeNode(5,val_3,null);
		TreeNode val_6 = new TreeNode(6);
		TreeNode root = new TreeNode(5,val_3,val_6);
		
		/*
		TreeNode val_5 = new TreeNode(5);
		TreeNode val_4 = new TreeNode(4,null,val_5);
		TreeNode val_2 = new TreeNode(2,null,val_4);
		TreeNode val_7 = new TreeNode(7);
		TreeNode val_6 = new TreeNode(6,null,val_7);
		TreeNode val_3 = new TreeNode(3,null,val_6);
		TreeNode root = new TreeNode(1,val_2,val_3);
		*/
		
		
		Solution1 so1 = new Solution1();
		System.out.println(so1.checkBalance(root));

	}

}