程序员面试题精选100题(60)-判断二叉树是不是平衡的

题目：输入一棵二叉树的根结点，判断该树是不是平衡二叉树。如果某二叉树中任意结点的左右子树的深度相差不超过1，那么它就是一棵平衡二叉树。例如下图中的二叉树就是一棵平衡二叉树：

#include <iostream>
using namespace std;

struct BTNode{
    BTNode *Left;
    BTNode *Right;
    int value; 
    BTNode(int val = 0)
        :value(val){}
};

BTNode* CreateTree(){
    int data;
    cin >> data;
    BTNode* root;

    if(-1 == data)
        return NULL;
    else{
        root = new BTNode(data);
        root->Left = CreateTree();
        root->Right = CreateTree();
    }
    return root;
}

bool IsBalancedBT(BTNode* pRoot,int *pDepth){
    if(!pRoot){
        *pDepth = 0;
        return true;
    }
    int left,right;
    if(IsBalancedBT(pRoot->Left,&left)
        &&IsBalancedBT(pRoot->Right,&right)){
        int diff = left - right;
        if(diff<=1 && diff>=-1){
            *pDepth = 1+(left>right?left:right);
            return true;
        }
    }
    return false;
}

bool IsBalanced(BTNode* root){
    int depth = 0;
    return IsBalancedBT(root,&depth);
}

void printTree(BTNode* root){
    if(!root) return;
    printTree(root->Left);
    cout << root->value << " ";
    printTree(root->Right);
    return;
}

int main(){
    BTNode* root = CreateTree();
    printTree(root);
    cout << endl;
    if(IsBalanced(root))
        cout << "The binary tree is balanced!" << endl;

    return 0;
}

参考文献： http://zhedahht.blog.163.com/blog/static/25411174201142733927831/