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.
问题：给定一个二叉树，判断其是否平衡。
当二叉树中所有节点的左右子树高度差不大于 1 时，这个二叉树视为平衡二叉树。
解题思路应该算一种分治思想。根据定义，递归地判断，实现判断。
const int unbalanceKids = -;
/**
* count the number of children of node and itself
*/
int height(TreeNode* node){
int heightL = ;
if (node->left != NULL){
heightL = height(node->left);
}
if(heightL == unbalanceKids){
return unbalanceKids;
}
int heightR = ;
if (node->right != NULL){
heightR = height(node->right);
}
if(heightR == unbalanceKids){
return unbalanceKids;
}
if (abs(heightL - heightR) > ){
return unbalanceKids;
}
return max(heightL, heightR) + ;
}
bool isBalanced(TreeNode* root) {
if (root == NULL){
return true;
}
int res = height(root);
if (res == unbalanceKids){
return false;
}
return true;
}