算法9:LeetCode_给定一个二叉树,判断它是否是高度平衡的二叉树。

我们在算法7中已经学习了二叉排序树,平衡二叉树的相关知识。既然我们知道平衡二叉树中,节点的左子树和右子树的高度差至多为1,那么我们就可以通过如下思路来进行判断。

1. 节点的左节点值小于根节点的值

2. 节点的右节点值大于根节点的值

3. 节点左右子树的高度差至多为1.

 

本体是Leetcode原题 https://leetcode.cn/problems/binary-tree-level-order-traversal-ii/,根据给定的事例我们知道,本题中判断一棵树是否是高度平衡,并不是说判断这棵树是否是平衡二叉树。此题中,我们只关注这棵树中的左节点和右节点的高度差,并不关注左节点和有节点的值谁大,这和平衡二叉树是有区别的。

package code.code_04;

/**
 * https://leetcode.cn/problems/balanced-binary-tree/
 * 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 。
 */
public class Code02_BalanceBinaryTree
{
    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;
        }
    }

    public static class Info {
        public boolean isBalanced;
        public int height;

        public Info(boolean i, int h) {
            isBalanced = i;
            height = h;
        }
    }

    public boolean isBalanced(TreeNode root) {
        return process(root).isBalanced;
    }

    public static Info process(TreeNode root) {
        //边界值
        if (root == null) {
            return new Info(true, 0);
        }

        Info leftInfo = process(root.left);
        Info rightInfo = process(root.right);
        //当前节点的高度
        int height = Math.max(leftInfo.height, rightInfo.height) + 1;
        boolean isBalanced = leftInfo.isBalanced && rightInfo.isBalanced
                && Math.abs(leftInfo.height - rightInfo.height) < 2;

        return new Info(isBalanced, height);
    }


    public static void main(String[] args) {
        Code02_BalanceBinaryTree tree = new Code02_BalanceBinaryTree();
        TreeNode node1 = tree.new TreeNode(3);
        TreeNode left2 = tree.new TreeNode(9);
        TreeNode righ2 = tree.new TreeNode(20);

        node1.left = left2;
        node1.right = righ2;

        TreeNode left3 = tree.new TreeNode(15);
        TreeNode righ3 = tree.new TreeNode(7);
        righ2.left = left3;
        righ2.right = righ3;

        System.out.println(tree.isBalanced(node1));
    }
}

  

 

posted @ 2022-12-10 22:59  街头小瘪三  阅读(21)  评论(0编辑  收藏  举报