二叉树平衡检查

  实现一个函数,检查二叉树是否平衡,平衡的定义如下,对于树中的任意一个结点,其两颗子树的高度差不超过1。给定指向树根结点的指针TreeNode* root,请返回一个bool,代表这棵树是否平衡。

import java.util.*;

/*
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;
    public TreeNode(int val) {
        this.val = val;
    }
}*/
public class Balance {
    public boolean isBalance(TreeNode root) {
        
        if(root == null){
            return true;
        }
        
        int lHeight = treeHeight(root.left);
        int rHeight = treeHeight(root.right);        
        if(lHeight - rHeight <= 1 && lHeight - rHeight >= -1){
            return isBalance(root.left) && isBalance(root.right);
        }
        
        return false;        
    }
    
    public int treeHeight(TreeNode root){
        if(root == null){
            return 0;
        }
        
        int lHeight = treeHeight(root.left);
        int rHeight = treeHeight(root.right);        
        return lHeight > rHeight ? lHeight + 1 : rHeight + 1;
    }
}
原文地址:https://www.cnblogs.com/liujinyao/p/4707149.html