LeetCode OJ：Balanced Binary Tree（平衡二叉树）

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.

就是去查看一棵树是不是平衡的，一开始对平衡二叉树的理解有错误，所以写错了 ，看了别人的解答之后更正过来了：

/**
 * 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) {
        int dep;
        checkBalance(root, dep);
    }
    bool checkBalance(TreeNode * root, int &dep)
    {
        if(root == NULL){
            dep = 0;
            return true;
        }
        int leftDep, rightDep;
        bool isLeftBal = checkBalance(root->left, leftDep);
        bool isRightBal = checkBalance(root->right, rightDep);

        dep = max(leftDep, rightDep) + 1;
        return isLeftBal && isRightBal && (abs(leftDep - rightDep) <= 1);
    }
};

pS：感觉这个不应该easy的题目啊  想的时候头还挺疼的。。

用java的时候用上面的方法去做总是无法成功，所以换了一种方法，这个一开始没有想到，是看别人写的，代码如下所示：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
public class Solution {
    public boolean isBalanced(TreeNode root) {
        if(root == null)
            return true; 
        if(root.left == null && root.right == null)
            return true;
        if(Math.abs(getDep(root.left) - getDep(root.right)) > 1)
            return false;
        return isBalanced(root.left) && isBalanced(root.right);
    }
    
    
    public int getDep(TreeNode node){
        if(node == null)
            return 0;
        else
            return 1 + Math.max(getDep(node.left), getDep(node.right));
    }
}