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.

判断一棵树是否为平衡二叉树。

思路:典型的递归,从第一个根节点开始,看左子树和右子树的深度差值是否小于等于1,然后再看由根节点的左孩子和右孩子构成的两课树是否都符合条件。

代码:

/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
private:
int maxdepth;
public:
void solve(TreeNode *root,int depth){
if(root==NULL)
return;
if(depth>maxdepth)
maxdepth=depth;
solve(root->left,depth+);
solve(root->right,depth+);
}
int getDepth(TreeNode *root) {
maxdepth=;
solve(root,);
return maxdepth;
}
bool isBalanced(TreeNode *root) {
if(root==NULL)
return true;
int l,r;
if(root->left!=NULL){
l=getDepth(root->left);
}else{
l=;
}
if(root->right!=NULL){
r=getDepth(root->right);
}else{
r=;
}
return abs(l-r)<=&&isBalanced(root->left)&&isBalanced(root->right);
}
};

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