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  • leetCode(20):Balanced binary tree 分类: leetCode 2015-06-22 08:52 190人阅读 评论(0) 收藏

    Given a binary tree, determine if it is height-balanced.

    For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

    方法一:求出左右子树的深度,如果差值大于1,则返回false,否则递归判断其左右孩子结点。但该方法每一次递归都要计算一次子结点的深度,出现重复计算。

    int maxDepth(TreeNode* root)
    {
    	if(NULL==root)
    		return 0;
    	int left=maxDepth(root->left);
    	int right=maxDepth(root->right);
    	
    	return 1+(left>right?left:right);
    }
    
    bool isBalanced(TreeNode* root)
    {
    	if(NULL==root)
    		return true;
    	int leftLength=maxDepth(root->left);
    	int rightLength=maxDepth(root->right);
    	int diff=leftLength-rightLength;
    	if(diff>1 || diff<-1)
    		return false;
    	return isBalanced(root->left) && isBalanced(root->right);
    }

    方法二:设置一个变量,用于记录深度,不必须重复计算。

    /**
     * Definition for a binary tree node.
     * struct TreeNode {
     *     int val;
     *     TreeNode *left;
     *     TreeNode *right;
     *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
     * };
     */
    class Solution {
    public:
       bool isBalanced(TreeNode* root,int *depth)
        {
        	if(NULL==root)
        	{
        		*depth=0;
        		return true;
        	}
        	int leftLength,rightLength;
        	if(isBalanced(root->left,&leftLength) && isBalanced(root->right,&rightLength))
        	{//若两棵子树都是平衡二叉树
        		int diff=leftLength-rightLength;
        		if(diff<=1 && diff>=-1)
        		{
        			*depth=1+(leftLength>rightLength?leftLength:rightLength);
        			return true;
        		}
        	}
        	return false;
        }
    
        bool isBalanced(TreeNode* root) {
            int depth;
        	return isBalanced(root,&depth);
        }
    };


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  • 原文地址:https://www.cnblogs.com/zclzqbx/p/4687095.html
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