判断二叉树是否平衡

//                      1
//              2             3
//        4                         5
//
//

class Solution {
public:
    int maxDepth(TreeNode* root){
        if(root==NULL)
            return 0;
 
        return 1+max(maxDepth(root->left),maxDepth(root->right));
    }
    bool danbian(TreeNode *root)
        {
        if(root==NULL)
            return 1;
        if(root->left!=NULL && root->right!=NULL)
            return 0;
        if(root->left==NULL)
            ;//danbian(root->right);
        if(root->right==NULL)
            ;//danbian(root->left);
        return 1;
    }
    int minDepth(TreeNode* root){
        
        if(root==NULL)
            return 0;

        

         if(root->left == NULL) return minDepth(root->right) + 1;
         if(root->right == NULL) return minDepth(root->left) + 1;

        int leftDepth = minDepth(root->left);
        int rightDepth = minDepth(root->right);
        return leftDepth < rightDepth ? (leftDepth + 1) : (rightDepth + 1);
    }
    bool IsBalanced_Solution(TreeNode* pRoot) {
        int max=maxDepth(pRoot);
        int min=minDepth(pRoot);
        if(danbian(pRoot))
            min=1;
        if(max-min>1)
            return 0;
        return 1;
        
        
    }
};

  

原文地址:https://www.cnblogs.com/laiqun/p/5894071.html