面试题55-题目二：平衡二叉树

最新推荐文章于 2025-06-23 08:00:00 发布

juaner1993

最新推荐文章于 2025-06-23 08:00:00 发布

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本文链接：https://blog.youkuaiyun.com/juaner1993/article/details/82760680

本文介绍了一种判断二叉树是否为平衡二叉树的方法。通过递归地计算每个节点的左右子树深度，并检查深度差是否超过1来实现。平衡二叉树是指任意节点的左右子树深度差不超过1的二叉树。

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/*

* 面试题55-题目二：平衡二叉树

* 题目：输入一棵二叉树，判断该二叉树是否是平衡二叉树。如果某二叉树中任意节点的左右子树的深度相差不超过1，那么它就是一棵平衡二叉树

* 思路：每个节点只遍历一次，采用后序遍历的方式遍历二叉树的每个节点

*/

public class No55IsBalanced_Solution {

    public static void main(String[] args) {

       No55IsBalanced_Solution n = new No55IsBalanced_Solution();

       TreeNode10 root = new TreeNode10(1);

       TreeNode10 p2 = new TreeNode10(2);

       TreeNode10 p3 = new TreeNode10(3);

       TreeNode10 p4 = new TreeNode10(4);

       TreeNode10 p5 = new TreeNode10(5);

       TreeNode10 p6 = new TreeNode10(6);

       TreeNode10 p7 = new TreeNode10(7);

       TreeNode10 p8 = new TreeNode10(8);

       TreeNode10 p9 = new TreeNode10(9);



       root.left = p2;

       root.right = p3;

       p2.left = p4;

       p2.right = p5;

       p4.left = p7;

       p4.right = p8;

       p7.left = p9;



       p3.right = p6;



       System.out.print(n.IsBalanced_Solution(root));

    }

    public boolean IsBalanced_Solution(TreeNode10 root) {

       if (root == null ) {

           return true;

       }



       int left = getDepth(root.left);

       int right = getDepth(root.right);



       int diff = left - right;



       if (diff >= -1 && diff <= 1) {

           return true;

       } else {

           return false;

       }

    }

    private int getDepth(TreeNode10 root) {

       if (root == null) {

           return 0;

       }



       int depth = 0;

       int leftNode = getDepth(root.left);

       int rightNode = getDepth(root.right);



       depth = leftNode > rightNode ? leftNode : rightNode;



       return depth + 1;

    }

}