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);
    }
};