[LeetCode] Balanced Binary Tree

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.

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思路：平衡二叉树就是和数的高度有关系，例外，如果有不是平衡二叉树的子树，那么返回-1，表示其不是平衡二叉树

/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool isBalanced(TreeNode *root)
{
if(root == NULL)
return true;
return treeHeight(root ) >= 0;
}

int treeHeight(TreeNode * root)
{
if(root == NULL)
return 0;
int lHeight = treeHeight(root->left);
int rHeight = treeHeight(root->right);
if(lHeight < 0 || rHeight < 0)
return -1;
if( abs(lHeight - rHeight) <= 1)
return max(lHeight, rHeight) + 1;
return -1;
}
};

另外一种思路：http://www.cnblogs.com/remlostime/archive/2012/10/27/2742987.html

class Solution {
public:
bool checkBalance(TreeNode *node, int &dep)
{
if (node == NULL)
{
dep = 0;
return true;
}

int leftDep, rightDep;
bool leftBalance = checkBalance(node->left, leftDep);
bool rightBalance = checkBalance(node->right, rightDep);

dep = max(leftDep, rightDep)+1;

return leftBalance && rightBalance && (abs(rightDep - leftDep) <= 1);
}

bool isBalanced(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int dep;
return checkBalance(root, dep);
}
};