Balanced Binary Tree

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.

思路：遍历这棵二叉树，每访问一个结点都要判断这个结点的子树是否是平衡的，如果平衡则继续访问结点；否则，可以判断这个二叉树是不平衡的。如果遍历完所有结点都没有找到不平衡的结点，说明该二叉树是平衡二叉树。

class Solution {
public:
    bool isBalanced(TreeNode *root) {
        stack<TreeNode *> s;//用于先序遍历
        TreeNode * pCur = root;
        int m, n;
        while(pCur || !s.empty())
        {
            m = height(pCur->left);
            n = height(pCur->right);
            if(m-n > 1 || m-n < -1)//找到不平衡点，结束
                return 0;
            s.push(pCur);  
            pCur = pCur->left;  
        //如果当前结点pCur为NULL且栈不空，则将栈顶结点出栈，  
        //同时置其右孩子为当前结点，循环判断，直至pCur不为空  
            while(!pCur && !s.empty())  
            {  
                pCur = s.top();  
                s.pop();  
                pCur = pCur->right;  
            }
        }
        return 1;//没找到不平衡点
    }
    
    int height(TreeNode *root)
    {
        if(!root)
            return -1;//空树高度定义为-1
        else if(!root->left && !root->right)//树只有一个根结点
            return 0;//只有一个结点的树，高度定义为0
        else
            return max(height(root->left), height(root->right)) + 1;
    }
    
    int max(int a, int b)//求两个值中的较大值
    {
        return a>b?a:b;
    }
};