Balanced Binary Tree(LeetCode)
2015-03-30 16:12
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题目: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.
示例代码如下:
class Solution {
public:
bool isBalanced(TreeNode *root) {
if(root==NULL)
return true;
int lheight=BinaryTreeHeight(root->left);//求左子树高度
int rheight=BinaryTreeHeight(root->right);//求右子树高度
if(abs(lheight-rheight)>1)//如果高度差绝对值大于1,返回false
return false;
else //判断左右子树是否是平衡二叉树
return isBalanced(root->left)&&isBalanced(root->right);
}
//求二叉树的高度
int BinaryTreeHeight(TreeNode* root){
if(root==NULL)
return 0;
int lheight=BinaryTreeHeight(root->left);
int rheight=BinaryTreeHeight(root->right);
return lheight>rheight ? lheight+1:rheight+1;
}
};
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.
示例代码如下:
class Solution {
public:
bool isBalanced(TreeNode *root) {
if(root==NULL)
return true;
int lheight=BinaryTreeHeight(root->left);//求左子树高度
int rheight=BinaryTreeHeight(root->right);//求右子树高度
if(abs(lheight-rheight)>1)//如果高度差绝对值大于1,返回false
return false;
else //判断左右子树是否是平衡二叉树
return isBalanced(root->left)&&isBalanced(root->right);
}
//求二叉树的高度
int BinaryTreeHeight(TreeNode* root){
if(root==NULL)
return 0;
int lheight=BinaryTreeHeight(root->left);
int rheight=BinaryTreeHeight(root->right);
return lheight>rheight ? lheight+1:rheight+1;
}
};
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