Lintcode：平衡二叉樹

問題：

給定一個二叉樹,確定它是高度平衡的。對於這個問題,一棵高度平衡的二叉樹的定義是：一棵二叉樹中每個節點的兩個子樹的深度相差不會超過1。 

樣例：

樣例  1:
	輸入: tree = {1,2,3}
	輸出: true
	
	樣例解釋:
	如下，是一個平衡的二叉樹。
		  1  
		 / \                
		2  3

	
樣例  2:
	輸入: tree = {3,9,20,#,#,15,7}
	輸出: true
	
	樣例解釋:
	如下，是一個平衡的二叉樹。
		  3  
		 / \                
		9  20                
		  /  \                
		 15   7 

python：

"""
Definition of TreeNode:
class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left, self.right = None, None
"""

class Solution:
    """
    @param root: The root of binary tree.
    @return: True if this Binary tree is Balanced, or false.
    """
    def deepth(self, root):
        if root == None:
            return 0
        return max(self.deepth(root.left), self.deepth(root.right))+1
    def isBalanced(self, root):
        # write your code here
        if root == None:
            return True
        if(abs(self.deepth(root.left) - self.deepth(root.right)) > 1):
            return False
        return (self.isBalanced(root.left) and self.isBalanced(root.right))

C++：

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */

class Solution {
public:
    /**
     * @param root: The root of binary tree.
     * @return: True if this Binary tree is Balanced, or false.
     */
    int deepth(TreeNode *root)
    {
        if(root == NULL)
        {
            return 0;
        }
        return (max(deepth(root->left),deepth(root->right)) + 1);
    }
    
    bool isBalanced(TreeNode * root) {
        // write your code here
        if(root == NULL)
        {
            return true;
        }
        if(abs(deepth(root->left)- deepth(root->right)) > 1)
        {
            return false;
        }
        return (isBalanced(root->left) && isBalanced(root->right));
    }
};