leetcode110. 平衡二叉树

only-qi2025-09-04 15:49

问题描述：

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。

示例 1:

给定二叉树 $3,9,20,null,null,15,7$

/ \

9 20

/ \

15 7

返回 true 。

示例 2:

给定二叉树 $1,2,2,3,3,null,null,4,4$

/ \

2 2

/ \

3 3

/ \

4 4

返回 false 。

思路：深度差小于2，返回true。

上代码，拿去即可运行：

java 复制代码

package onlyqi.daydayupgo06.leetcode;
 
public class TreeNode {
 
    private Integer value;
    private TreeNode left;
    private TreeNode right;
 
 
    public TreeNode() {
    }
 
    public TreeNode(Integer value) {
        this.value=value;
    }
 
    public TreeNode(Integer value, TreeNode left, TreeNode right) {
        this.value = value;
        this.left = left;
        this.right = right;
    }
 
    public Integer getValue() {
        return value;
    }
 
    public void setValue(Integer value) {
        this.value = value;
    }
 
    public TreeNode getLeft() {
        return left;
    }
 
    public void setLeft(TreeNode left) {
        this.left = left;
    }
 
    public TreeNode getRight() {
        return right;
    }
 
    public void setRight(TreeNode right) {
        this.right = right;
    }
}

java 复制代码

public class Tree {


    public static void main(String[] args) {
        TreeNode treeNode1 = new TreeNode(1);
        TreeNode treeNode2 = new TreeNode(2);
        TreeNode treeNode3 = new TreeNode(3);
        TreeNode treeNode4 = new TreeNode(4);
        TreeNode treeNode5 = new TreeNode(5);
        TreeNode treeNode6 = new TreeNode(6);
        TreeNode treeNode7 = new TreeNode(7);
        treeNode2.setLeft(treeNode4);
        treeNode2.setRight(treeNode5);
        treeNode1.setLeft(treeNode2);
        treeNode1.setRight(treeNode3);
        treeNode3.setLeft(treeNode6);

        System.out.println(isBalanced(treeNode1));

    }


    public static boolean isBalanced(TreeNode root) {
        return depth(root) != -1;
    }
    private static int depth(TreeNode root) {
        if (root == null) return 0;
        int left = depth(root.getLeft());
        if(left == -1) return -1;
        int right = depth(root.getRight());
        if(right == -1) return -1;
        return Math.abs(left - right) < 2 ? Math.max(left, right) + 1 : -1;
    }
}

运行结果：

我要刷300道算法题，第143道 。尽快刷到200，每天搞一道。