平衡二叉樹

/*
問題描述：平衡二叉樹，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。
來源：LintCode
作者：syt

日期：2017-11-10

思路：計算每個結點的左子樹和右子樹的深度，並判斷是否是平衡二叉樹

*/

#include <iostream>

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

#include "d_tree.h"
#include <queue>
using namespace std;

/*
* @param root: The root of binary tree.
* @return: True if this Binary tree is Balanced, or false.
*/
int calDepth(TreeNode* root)
{
	if (root == NULL)
		return 0;
	int left = calDepth(root->left);
	int right = calDepth(root->right);

	int max = 0;
	if (left > right)
		max = left + 1;
	else
		max = right + 1;
	return max;
}

bool isBalanced(TreeNode * root) {
	// write your code here
	if (root == NULL)
		return true;
	queue<TreeNode *> q;
	TreeNode* p;
	q.push(root);
	while (!q.empty())
	{
		p = q.front();
		q.pop();

		int left = calDepth(p->left);
		int right = calDepth(p->right);
		int res = left - right;

		if (res > 1 || res < -1)
			return false;

		if (p->left != NULL)
			q.push(p->left);
		if (p->right != NULL)
			q.push(p->right);
	}

	return true;
}


TreeNode *buildTree()
{
	TreeNode *root, *b, *c, *d, *e, *f, *g, *h;
	h = new TreeNode(8);
	g = new TreeNode(7);
	f = new TreeNode(6);
	e = new TreeNode(5);
	d = new TreeNode(4, g, h);	
	c = new TreeNode(3, e, f);
	b = new TreeNode(2, d, (TreeNode *)NULL);
	root = new TreeNode(1, b, c);

	return root;
}

void main()
{

	TreeNode *root = buildTree();
	cout << isBalanced(root) << endl;

}