【遍历二叉树】10判断二叉树是否平衡【Balanced Binary Tree】

平衡的二叉树的定义都是递归的定义，所以，用递归来解决问题，还是挺容易的额。

本质上是递归的遍历二叉树。

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给定一个二叉树，判定他是不是高度平衡的二叉树。

对于这个问题，每个节点的两个子树的深度不会相差超过1，那么这样的二叉树就是一个平衡的二叉树

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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.

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test.cpp：

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#include <iostream>
#include <cstdio>
#include <stack>
#include <vector>
#include "BinaryTree.h"

using namespace std;

/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
bool balance(TreeNode *root, int &depth)
{
    if(root == NULL)
    {
        depth = 0;
        return true;
    }
    int leftdepth = 0;
    bool left = balance(root->left, leftdepth);

int rightdepth = 0;
bool right = balance(root->right, rightdepth);

depth = leftdepth > rightdepth ? leftdepth + 1 : rightdepth + 1;
int gap = leftdepth - rightdepth;

return left && right && (-1 <= gap && gap <= 1);
}

bool isBalanced(TreeNode *root)
{

int depth = 0;
return balance(root, depth);
}

// 树中结点含有分叉，
//                  6
//              /       \
//             7         2
//           /   \
//          1     4
//               / \
//              3   5
int main()
{
    TreeNode *pNodeA1 = CreateBinaryTreeNode(6);
    TreeNode *pNodeA2 = CreateBinaryTreeNode(7);
    TreeNode *pNodeA3 = CreateBinaryTreeNode(2);
    TreeNode *pNodeA4 = CreateBinaryTreeNode(1);
    TreeNode *pNodeA5 = CreateBinaryTreeNode(4);
    TreeNode *pNodeA6 = CreateBinaryTreeNode(3);
    TreeNode *pNodeA7 = CreateBinaryTreeNode(5);

ConnectTreeNodes(pNodeA1, pNodeA2, pNodeA3);
ConnectTreeNodes(pNodeA2, pNodeA4, pNodeA5);
ConnectTreeNodes(pNodeA5, pNodeA6, pNodeA7);

// 树中结点含有分叉，
    //                  1
    //              /       \
    //             2         2
    //           /   \       / \
    //          3     4     4   3

TreeNode *pNodeB1 = CreateBinaryTreeNode(1);
    TreeNode *pNodeB2 = CreateBinaryTreeNode(2);
    TreeNode *pNodeB3 = CreateBinaryTreeNode(2);
    TreeNode *pNodeB4 = CreateBinaryTreeNode(3);
    TreeNode *pNodeB5 = CreateBinaryTreeNode(4);
    TreeNode *pNodeB6 = CreateBinaryTreeNode(4);
    TreeNode *pNodeB7 = CreateBinaryTreeNode(3);

ConnectTreeNodes(pNodeB1, pNodeB2, pNodeB3);
ConnectTreeNodes(pNodeB2, pNodeB4, pNodeB5);
ConnectTreeNodes(pNodeB3, pNodeB6, pNodeB7);

bool ans = isBalanced(pNodeA1);

if (ans == true)
    {
        cout << "Balanced!" << endl;
    }
    else
    {
        cout << "Not Balanced!" << endl;
    }

bool ans1 = isBalanced(pNodeB1);

if (ans1 == true)
    {
        cout << "Balanced!" << endl;
    }
    else
    {
        cout << "Not Balanced!" << endl;
    }
    DestroyTree(pNodeA1);
    DestroyTree(pNodeB1);
    return 0;
}

结果输出：

Not Balanced!

Balanced!

BinaryTree.h:

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#ifndef _BINARY_TREE_H_
#define _BINARY_TREE_H_

struct TreeNode
{
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

TreeNode *CreateBinaryTreeNode(int value);
void ConnectTreeNodes(TreeNode *pParent,
TreeNode *pLeft, TreeNode *pRight);
void PrintTreeNode(TreeNode *pNode);
void PrintTree(TreeNode *pRoot);
void DestroyTree(TreeNode *pRoot);

#endif /*_BINARY_TREE_H_*/

BinaryTree.cpp:

#include <iostream>
#include <cstdio>
#include "BinaryTree.h"

using namespace std;

/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/

//创建结点
TreeNode *CreateBinaryTreeNode(int value)
{
TreeNode *pNode = new TreeNode(value);

return pNode;
}

//连接结点
void ConnectTreeNodes(TreeNode *pParent, TreeNode *pLeft, TreeNode *pRight)
{
    if(pParent != NULL)
    {
        pParent->left = pLeft;
        pParent->right = pRight;
    }
}

//打印节点内容以及左右子结点内容
void PrintTreeNode(TreeNode *pNode)
{
    if(pNode != NULL)
    {
        printf("value of this node is: %d\n", pNode->val);

if(pNode->left != NULL)
            printf("value of its left child is: %d.\n", pNode->left->val);
        else
            printf("left child is null.\n");

if(pNode->right != NULL)
            printf("value of its right child is: %d.\n", pNode->right->val);
        else
            printf("right child is null.\n");
    }
    else
    {
        printf("this node is null.\n");
    }

printf("\n");
}

//前序遍历递归方法打印结点内容
void PrintTree(TreeNode *pRoot)
{
PrintTreeNode(pRoot);

if(pRoot != NULL)
    {
        if(pRoot->left != NULL)
            PrintTree(pRoot->left);

if(pRoot->right != NULL)
PrintTree(pRoot->right);
}
}

void DestroyTree(TreeNode *pRoot)
{
    if(pRoot != NULL)
    {
        TreeNode *pLeft = pRoot->left;
        TreeNode *pRight = pRoot->right;

delete pRoot;
pRoot = NULL;

DestroyTree(pLeft);
DestroyTree(pRight);
}
}

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【遍历二叉树】10判断二叉树是否平衡【Balanced Binary Tree】

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