【面试题039】二叉树的深度

【面试题039】二叉树的深度

题目一：

输入一棵二叉树的根结点，求该树的深度。从根结点到叶结点依次经过的结点（含根，叶结点）形成树的一条路径，最长路径的长度为树的深度。

二叉树结点的结构如下：

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struct BinaryTreeNode
{
    int m_nValue;
    BinaryTreeNode *m_pLeft;
    BinaryTreeNode *m_pRight;
}

思路一：

如果一个树只有一个结点，那么这个树的深度为1，

如果根结点只有左子树而没有右子树，那么树的深度应该是其左子树的深度加1；

同样如果根结点只有右子树而没有左子树，那么树的深度应该是其右子树的深度加1；

如果基友左子树又有右子树，那么该树的深度就是其左、右子树深度的最大值再加1；

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

using namespace std;

int TreeDepth(BinaryTreeNode *pRoot)
{
    if(pRoot == NULL)
        return 0;

    int nLeft = TreeDepth(pRoot->m_pLeft);
    int nRight = TreeDepth(pRoot->m_pRight);

    return (nLeft > nRight) ? (nLeft + 1) : (nRight + 1);
}
//            1
//         /
//        2        3
//       /
//      4  5         6
//        /
//       7
int main()
{
    BinaryTreeNode *pNode1 = CreateBinaryTreeNode(1);
    BinaryTreeNode *pNode2 = CreateBinaryTreeNode(2);
    BinaryTreeNode *pNode3 = CreateBinaryTreeNode(3);
    BinaryTreeNode *pNode4 = CreateBinaryTreeNode(4);
    BinaryTreeNode *pNode5 = CreateBinaryTreeNode(5);
    BinaryTreeNode *pNode6 = CreateBinaryTreeNode(6);
    BinaryTreeNode *pNode7 = CreateBinaryTreeNode(7);

    ConnectTreeNodes(pNode1, pNode2, pNode3);
    ConnectTreeNodes(pNode2, pNode4, pNode5);
    ConnectTreeNodes(pNode3, NULL, pNode6);
    ConnectTreeNodes(pNode5, pNode7, NULL);

    cout << TreeDepth(pNode1) << endl;

    DestroyTree(pNode1);
    return 0;
}

BinaryTree.h

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#ifndef _BINARY_TREE_H_
#define _BINARY_TREE_H_

struct BinaryTreeNode
{
    int                    m_nValue;
    BinaryTreeNode        *m_pLeft;
    BinaryTreeNode        *m_pRight;
};

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

#endif //_BINARY_TREE_H_

BinaryTree.cpp

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

BinaryTreeNode *CreateBinaryTreeNode(int value)
{
    BinaryTreeNode *pNode = new BinaryTreeNode();
    pNode->m_nValue = value;
    pNode->m_pLeft = NULL;
    pNode->m_pRight = NULL;

    return pNode;
}

void ConnectTreeNodes(BinaryTreeNode *pParent,
                      BinaryTreeNode *pLeft, BinaryTreeNode *pRight)
{
    if(pParent != NULL)
    {
        pParent->m_pLeft = pLeft;
        pParent->m_pRight = pRight;
    }
}

void PrintTreeNode(BinaryTreeNode *pNode)
{
    if(pNode != NULL)
    {
        printf("value of this node is: %d ", pNode->m_nValue);

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

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

    printf(" ");
}

void PrintTree(BinaryTreeNode *pRoot)
{
    PrintTreeNode(pRoot);

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

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

void DestroyTree(BinaryTreeNode *pRoot)
{
    if(pRoot != NULL)
    {
        BinaryTreeNode *pLeft = pRoot->m_pLeft;
        BinaryTreeNode *pRight = pRoot->m_pRight;

        delete pRoot;
        pRoot = NULL;

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

题目二：

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。

思路一：

根据深度来做判断，但是结点重复遍历了很多次，效率不高。

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bool IsBalanced(BinaryTreeNode *pRoot)
{
    if (pRoot == NULL)
    {
        return true;
    }
    int left = TreeDepth(pRoot->m_pLeft);
    int right = TreeDepth(pRoot->m_pRight);
    int diff = left - right;
    if (diff > 1 || diff < -1)
    {
        return false;
    }
    return IsBalanced(pRoot->m_pLeft)
           && IsBalanced(pRoot->m_pRight);
}

思路二：

记住，每个结点都遍历一遍的解法才是高效的解法。

用后序遍历的方式遍历二叉树的每一个结点，在遍历到一个结点之前我们就已经遍历了它的左右子树。

只要在遍历每个结点的时候记录它的深度，我们就可以一边遍历，一边判断是不是平衡的。

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bool IsBalanced(BinaryTreeNode *pRoot, int *pDepth)
{
    if (pRoot == NULL)
    {
        *pDepth = 0;
        return true;
    }
    int left, right;
    if (IsBalanced(pRoot->m_pLeft, &left)
            && IsBalanced(pRoot->m_pRight, &right))
    {
        int diff = left - right;
        if (diff <= 1 && diff >= -1)
        {
            *pDepth = 1 + (left > right ? left : right);
            return true;
        }
    }
    return false;
}

bool IsBalanced(BinaryTreeNode *pRoot)
{
    int depth = 0;
    return IsBalanced(pRoot, &depth);
}