二叉树

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一、二叉树的性质

1、第K层的二叉树节点个数最多为2^(k - 1)个

2、k层的二叉树节点个数最多有2^(k) - 1个

3、度数为2的节点个数n2等于叶子节点个数n0减，即n2 + 1 = n0

4、第i个节点的左孩子为2i，右孩子为（2i + 1），根节点从1开始

5、具有N个节点的二叉树的深度为log2(N)的下界 + 1

二、树的表示方法
1、二叉链表表示法

typedef struct TreeNode{    int date;    struct TreeNode *lNode, *rNode;}TreeNode;

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2、三叉链表表示法

typedef struct TreeNode{    int date;    struct TreeNode *lNode, *rNode， *parent;}TreeNode;

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3、双亲链表表示法

typedef struct BTNode{    int date;    int parentPosition;      //根节点的下标    int LRtag;        //左右孩子的标志}BTNode;typedef struct BTree{    BTNode node[100];    int nodenum;    int root;}BTree;int main(){    BTree tree;    tree.root = 0;    tree.node[0].date = 0;    tree.node[1].date = 1;    tree.node[2].date = 2;    tree.node[3].date = 3;    tree.node[4].date = 4;    tree.node[1].parentPostion = 0;    tree.node[1].LRtag = 1;    tree.node[2].parentPostion = 0;    tree.node[2].LRtag = 2;    tree.node[3].parentPostion = 1;    tree.node[3].LRtag = 1;    tree.node[4].parentPostion = 2;    tree.node[4].LRtag = 2; }

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三、二叉树的遍历

1、二叉树递归遍历的本质

（1）先序遍历

typedef struct TreeNode{    int date;    struct TreeNode *lNode, *rNode;}TreeNode;void preoder(TreeNode* t){    if(t == NULL)        return;    printf("%d\n", t->date);    preoder(t->lNode);    preoder(t->rNode);}

（2）中序遍历

typedef struct TreeNode{    int date;    struct TreeNode *lNode, *rNode;}TreeNode;void inoder(TreeNode* t){    if(t == NULL)        return;    inoder(t->lNode);    printf("%d\n", t->date);    inoder(t->rNode);}

（1）后序遍历

typedef struct TreeNode{    int date;    struct TreeNode *lNode, *rNode;}TreeNode;void postoder(TreeNode* t){    if(t == NULL)        return;    postoder(t->lNode);    postoder(t->rNode);    printf("%d\n", t->date);}

三种递归遍历的本质是一样的，如果把printf的输出语句去掉，代码都是一样的，在二叉树的遍历过程中，所有的节点都会经过三次，如果每个节点经过第一次就访问，即为先序遍历，如果每个节点经过第二次再访问，即为中序遍历，如果每个节点经过第三次再访问，即为后序遍历；所以说递归遍历二叉树的本质就是遍历都是先左孩子再右孩子，只不过访问节点的时机不同罢了。

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下面的二叉树均是以这样的二叉树结构体实现的：

typedef struct _tag_BTree {    char val;    struct _tag_BTree* left, *right;}BTree;

2、非递归遍历

（1）先序遍历

void unrecpreoder(BTree* TreeNodeA) {    BTree* p = TreeNodeA;    stack<BTree*> st;    while (p || !st.empty()) {        while (p) {            st.push(p);            cout << p->val;            p = p->left;        }        p = st.top();        st.pop();        p = p->right;    }    return;}

2、中序遍历

void unrecinoder(BTree* TreeNodeA) {
BTree* p = TreeNodeA;
stack

void unrecpostoder(BTree* TreeNodeA) {    BTree* p = TreeNodeA;    BTree* q = NULL;    stack<BTree*> st;    while (p || !st.empty()) {        while (p) {            st.push(p);            p = p->left;        }        p = st.top();        if (p->right == NULL || p->right == q) {            q = st.top();            cout << q->val;            st.pop();            p = NULL;        }        else            p = p->right;    }    return;}

四、求所有节点个数

int getleafnum(BTree* TreeNode, int * num) {    int leafnum = 0;    if (TreeNode == NULL)        return 0;    leafnum++;    int leftnum = getleafnum(TreeNode->left, num);    int rightnum = getleafnum(TreeNode->right, num);    leafnum += leftnum + rightnum;    *num = leafnum;    return leafnum;}

五、求二叉树的深度

int getdepth(BTree* TreeNode, int *depth) {    int treedepth = 0;    if (TreeNode == NULL)        return 0;    int leftdepth = getdepth(TreeNode->left, depth);    int rightdepth = getdepth(TreeNode->right, depth);    treedepth = max(leftdepth, rightdepth) + 1;    *depth = treedepth;    return treedepth;}

六、二叉树的拷贝

BTree* TreeCopy(BTree* TreeNode) {    if (TreeNode == NULL)        return NULL;    BTree* newnode = (BTree*)malloc(sizeof(BTree));    newnode->val = TreeNode->val;    newnode->left = TreeCopy(TreeNode->left);    newnode->right = TreeCopy(TreeNode->right);    return newnode;}

未完待续……

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