Balanced Binary Tree

這個麻煩在，對以一個結點爲根的樹，左右樹都Balance，左右樹的差小於1同時滿足才行。第一種，每一個算Balance都要算到下面的height，然後複雜度O（N^2）

/**
 * Definition for a binary tree node.
 * 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;
         if(root->left==NULL)
         {
             if(height(root->right)>0)
                return false;
              
              return true;
              
         }
         
         if(root->right==NULL)
         {
             if(height(root->left)>0)
                return false;
                
             return true;
           
         }
         
         
         if(isBalanced(root->left)&&isBalanced(root->right)&&(abs(height(root->left)-height(root->right))<=1))
         {
             return true;
         }
         
        
         return false;
    }
    
    int height(TreeNode * root)
    {
        if(root==NULL)
            return -1;
         
        int a=height(root->left);
        int b=height(root->right);
        return (a>b)?(a+1):(b+1);
    }
    
    
};

第二種，基於DFS，這種用一個-1做標識，最後不是-1的就是banlance，過程中只要有不平衡就會標記-1。這樣複雜度是O（n）

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isBalanced(TreeNode* root) {
        return dfsHeight(root)!=-1;
    }
    
    
    int dfsHeight(TreeNode * root)
    {
        if(root==NULL) return 0;
        
        int leftHeight=dfsHeight(root->left);
        if(leftHeight==-1) return -1;
        int rightHeight=dfsHeight(root->right);
        if(rightHeight==-1) return -1;
        
        if(abs(leftHeight-rightHeight)>1)
        return -1;
        return max(leftHeight,rightHeight)+1;
    }
};