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tree.tex
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tree.tex
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\section{Tree}
\begin{frame}[fragile]{树和二叉树}
\begin{columns}[T]
\column{0.5\textwidth}
树型结构是结点之间有分支,并且具有层次关系的结构,类似于自然界中的树。树有很多
应用,比如Unix等操作系统中的目录结构。
\column{0.4\textwidth}
\includegraphics[width=0.9\textwidth]{figs/linux-tree.png}
\end{columns}
\end{frame}
\begin{frame}[fragile]
\frametitle{例子}
\begin{forest}
[CEO, for tree={rectangle, minimum width=2cm}, fill=red!10
[CFO [财务人员] ]
[CTO [工程师] ]
[CMO [销售人员] ]
]
\node at (current bounding box.south)
[below=1ex,draw,cloud,aspect=6,cloud puffs=30]
{\emph{Simple Company Hierarchy}};
\end{forest}
\end{frame}
\begin{frame}[fragile, plain]
\scalebox{0.7}{
\begin{forest}
[学院, for tree={draw=none, rectangle, minimum width=1cm}, fill=red!10, circle
[社会学部, grow=west [信息资源管理学院, fill=red!10] [新闻学院] [农业与农村发展学院] [社会与人口学院]
[公共管理学院] [教育学院]]
[$\cdots$]
[ 人文学部,grow=east [哲学院] [文学院] [历史学院] [国学院] [艺术学院] [外国语学院] [清史研究所]]
]
\node at (current bounding box.south)
[below=1ex,draw,cloud,aspect=6,cloud puffs=30]
{\emph{人民大学学院设置}};
\end{forest}
}
\end{frame}
\begin{frame}[fragile]{内容}
\begin{easylist} \easyitem
& 树的基本术语
& 二叉树
& 遍历二叉树与线索二叉树
& 树和森林
& 哈夫曼树
\end{easylist}
\end{frame}
\subsection{基本术语}
\begin{frame}[fragile]
\frametitle{树(TREE)}树(Tree)是$n(n \geq 0)$个结点的有限集$T$。 $T$为空时称为空
树。当$n>0$时,树有且仅有一个特定的称为根(Root)的结点,其余结点可分为$m(m \geq
0)$个互不相交的子集$T_1, T_2, \cdots, T_m$,其中每个子集又是一棵树,称为子
树(Subtree)。
\begin{columns}[T]
\column{0.55\textwidth}
\begin{easylist}
& 各子树是互不相交的集合。
& 除根结点,其它结点有唯一前驱。
& 一个结点可以有零个或多个后继。
\end{easylist}
\column{0.4\textwidth}
\begin{forest}
[R, for tree={color=white,fill=black}, fill=red!85
[A [C] [D] [E]]
[B [F]]
]
\end{forest}
\end{columns}
\end{frame}
\begin{frame}[fragile]
\frametitle{判断哪些是树结构}
\includegraphics[width=0.3\textwidth]{dot/tree-judge1.pdf} ~~~~~
\pause
\includegraphics[width=0.4\textwidth]{dot/tree-judge2.pdf}
\end{frame}
\begin{frame}[fragile]
\frametitle{判断哪些是树结构}
\includegraphics[width=0.35\textwidth]{dot/tree-judge3.pdf} ~~~~~
\pause
\includegraphics[width=0.4\textwidth]{dot/tree-judge4.pdf}
\end{frame}
\begin{frame}[fragile]
\frametitle{树的表示形式}
\includegraphics[width=0.36\textwidth]{dot/tree-represent1.pdf} \pause
\scalebox{0.75}{
\begin{tikzpicture}[b/.style={fill=black!50},n/.style={minimum width=1cm}]
\draw node[n] (a) {A} node[b, right=0 of a, minimum width=5cm, fill=red!50]{};
\draw node[minimum width=0.5cm, below=0.1 of a](bh){}
node[n, right=0 of bh] (b) {B} node[b,fill=blue!50, minimum width=4.3cm,right=0 of b]{};
\draw node[minimum width=2cm, below=0.2 of bh](dh){}
node[n, right=0 of dh] (d) {D} node[b, minimum width=3.5cm,right=0 of d]{};
\draw node[minimum width=3.5cm, below=0.2 of dh](ih){}
node[n, right=0 of ih] (i) {I} node[b, fill=green!50, minimum width=2.8cm,right=0 of i]{};
\draw node[minimum width=3.5cm, below=0.2 of ih](jh){}
node[n, right=0 of jh] (j) {J} node[b, fill=green!50, minimum width=2.8cm,right=0 of j]{};
\draw node[minimum width=2cm, below=1.2 of dh](eh){}
node[n, right=0 of eh] (e) {E} node[b, minimum width=3.5cm,right=0 of e]{};
\draw node[minimum width=2cm, below=1.8 of dh](fh){}
node[n, right=0 of fh] (f) {F} node[b, minimum width=3.5cm,right=0 of f]{};
\draw node[minimum width=0.5cm, below=3.2 of a](ch){}
node[n, right=0 of ch] (c) {C} node[b, fill=blue!50, minimum width=4.3cm,right=0 of c]{};
\draw node[minimum width=2cm, below=2.8 of dh](gh){}
node[n, right=0 of gh] (g) {G} node[b, minimum width=3.5cm,right=0 of g]{};
\draw node[minimum width=2cm, below=3.2 of dh](hh){}
node[n, right=0 of hh] (h) {H} node[b, minimum width=3.5cm,right=0 of h]{};
\draw node[below=0.1 of h] {凹入表表示法};
\end{tikzpicture}
} \pause
\begin{columns}[T]
\begin{column}{0.4\textwidth}
\centering
\vspace{2cm}
(A(B(D(I,J),E, F),C(G,H)))
广义表表示 \\
\pause
\end{column}
\begin{column}{0.5\textwidth}
\scalebox{0.65}{
\begin{tikzpicture}[n/.style={ellipse,draw}]
\draw node[n, minimum width=7.8cm, minimum height=3.5cm, fill=red!5]{}
node[n, minimum width=4.5cm, minimum height=2.5cm, xshift=-1.2cm, fill=blue!5]{}
node[n, minimum width=2.5cm, minimum height=1.8cm, xshift=2.5cm, fill=blue!5]{}
node[n, minimum width=2cm, minimum height=1.8cm, xshift=-2cm, fill=yellow!5]{}
node[n, circle,xshift=-2.5cm, fill=green!5]{I}
node[n, circle,xshift=-1.7cm, fill=green!5]{J}
node[n, circle,xshift=-0.5cm, fill=yellow!5]{E}
node[n, circle,xshift=0.4cm, fill=yellow!5]{F}
node[n, circle,xshift=2cm, fill=yellow!5]{G}
node[n, circle,xshift=3cm, fill=yellow!5]{H}
node[yshift=1.35cm]{A}
node[xshift=-0.8cm,yshift=0.8cm]{B}
node[xshift=2.5cm, yshift=0.6cm]{C}
node[xshift=-2cm,yshift=0.6cm]{D};
\draw node[yshift=-2.2cm] {嵌套集合表示};
\end{tikzpicture}
}
\end{column}
\end{columns}
\end{frame}
\begin{frame}[fragile]
\frametitle{基本术语}
\begin{columns}[t]
\begin{column}{0.5\textwidth}
\begin{itemize}
\item 树(tree)
\item 子树(sub-tree)
\item 结点(node)
\item 结点的度(degree)
\item 叶子(leaf)
\item 孩子(child)
\item 父亲(parents)
\item 兄弟(sibling)
\item 祖先
\item 子孙
\end{itemize}
\end{column}
\begin{column}{0.5\textwidth}
\begin{itemize}
\item 树的度(degree)
\item 结点的层次(level)
\item 树的深度(depth)
\item 有序树
\item 无序树
\item 森林
\end{itemize}
\scalebox{0.8} {
\begin{forest}
[R, for tree={color=white,fill=black}, fill=red!85
[A [C] [D]]
[B [E]]
]
\end{forest}
}
\end{column}
\end{columns}
\end{frame}
\begin{frame}[fragile, plain]
~
\end{frame}
\subsection{二叉树}
\begin{frame}[fragile]
\frametitle{二叉树(Binary Tree)}
\begin{itemize}
\item 二叉树是一种树型结构,它的每个结点至多只有两个子树,分别称为左子树和右子树。
二叉树是有序树。
\item 二叉树是$n(n \geq 0)$个结点构成的有限集合。二叉树或为空,或是由一个根结点及
两棵互不相交的左右子树组成,并且左右子树都是二叉树。
\item 在二叉树中要区分左子树和右子树,即使只有一棵子树。这是二叉树与树的最主要的
差别。
\end{itemize}
二叉树的一个重要应用是在查找中的应用。当然,它还 有许多与搜索无关的重要应用,比如
在编译器的设计领域。
\end{frame}
\begin{frame}[fragile]
\frametitle{二叉树的五种形态}
\begin{enumerate}
\item 空二叉树;
\item 只有根结点(左右子树都为空);
\item 只有左子树(右子树为空);
\item 只有右子树(左子树为空);
\item 左右子树均不空。
\end{enumerate}
\scalebox{0.8}{
\begin{tikzpicture}[n/.style={draw=black!80, thick,fill=black!80, circle, minimum size=0.6cm},
t/.style={draw=black!80, ellipse, minimum height=2cm,minimum width=1cm, fill=black!20}]
\draw node[n,dotted, fill=yellow!1] (t1) at (0,0) {};
\draw[draw=red] (0.5,0.5) -- (-0.5,-0.5);
\draw node[n] (t2) at (2,0) {};
\draw node[n] (t3) at (4.5,0) {} node[t, below left=of t3,xshift=0.8cm, rotate=-30] (t31) {};
\draw[draw] (t3) -- (t31);
\draw node[n] (t4) at (6.5,0) {} node[t, below right=of t4,xshift=-0.8cm, rotate=30] (t41) {};
\draw[draw] (t4) -- (t41);
\draw node[n] (t5) at (11,0) {} node[t, below left=of t5,xshift=0.8cm, rotate=-30] (t51) {} node[t, below right=of t5,xshift=-0.8cm, rotate=30] (t52) {};
\draw[draw] (t5) -- (t51);
\draw[draw] (t5) -- (t52);
\end{tikzpicture}
}
\end{frame}
\begin{frame}[fragile]
\frametitle{请观察二叉树, 并回答下列问题}
\scalebox{0.7} {
\begin{forest}
[A, name=A
[B, name=B [D [H] [I]] [E [J] [K]]]
[C, name=C [F [L] [M]] [G [N] [O]]]
];
\draw[draw, dotted, thick] (B.north) ++(-2cm, 0.2cm)-- ++(9cm, 0cm)
node[above, right, yshift=0.5cm]{1};
\draw[draw, dotted, thick] (B.south) ++(-2cm, -0.2cm)-- ++(9cm, 0cm)
node[above, right, yshift=0.5cm]{2};
\draw[draw, dotted, thick] (B.south) ++(-2cm, -1.6cm)-- ++(9cm, 0cm)
node[above, right, yshift=0.5cm]{3} node[below, right, yshift=-0.5cm] {4};
\end{forest}
}
\begin{enumerate}
\item 二叉树的第$i$层最多有多少个结点?
\item 二叉树深度为k,则它最多有多少个结点?
\item 二叉树有n个节点,请问它最小深度是几?
\item 二叉树叶子的数目和度为2的节点的数目是否相等?如果不等,又是什么关系?
\end{enumerate}
\end{frame}
\begin{frame}[fragile]
\frametitle{二叉树的性质}
\begin{itemize}
\item 性质1: 二叉树的第$i$层至多有$2^{i-1}$个结点。
\item 性质2: 深度为$k$的二叉树至多有$2^k-1$个结点($k \geq 1$)。
\item \color{red} 性质3: 二叉树中终端结点数为$n_0$,度为2的结点数为$n_2$,则有$n_0 = n_2 + 1$
(试证明)
\end{itemize}
\end{frame}
\begin{frame}[fragile]
\frametitle{二叉树的性质}
二叉树中终端结点数为$n_0$,度为2的结点数为$n_2$,则有$n_0 = n_2 + 1$
\begin{columns}[T,c]
\begin{column}{0.6\linewidth}
\begin{itemize}
\item 设二叉树中度为1的结点数为$n_1$, 二叉树中总结点数为$N$,则有:
$N = n_0 + n_1 + n_2$
\item 再考虑二叉树中的分支数(每个节点有唯一一个入的分支,根节点除外;再考虑出
的分支数量), 则有:
$N - 1 =n_1 + 2 \times n_2$
\item 整理可得:
$n_0 = n_2 + 1$
\end{itemize}
\end{column}
\begin{column}{0.35\linewidth}
\begin{forest}
[{}, name=root,for tree={color=white,fill=black}
[{}, name=n21, grow=-55 [{}, name=n31 [{~}] [{~}]]]
[{}, name=n22, grow=235 [{}, name=n32,[{~}, draw=none,fill=none, no edge] [{~}]]]
]
\end{forest}
\end{column}
\end{columns}
\end{frame}
\begin{frame}[fragile]
\frametitle{完全二叉树}
\begin{columns}[T]
\column{0.55\textwidth}
\begin{forest}
[1, for tree={minimum width=1.8em, minimum height=1.8em}
[2 [4 [8] [9]] [5 [10] [11]]]
[3 [6 [12] [{}, no edge,draw=none]] [7]]
]
\end{forest}
\pause
\column{0.4\textwidth}
是否完全二叉树?
\begin{forest}
[1
[2 [4] [5]]
[3 [{}, no edge, draw=none] [7]]
]
\end{forest}
\pause
\begin{forest}
[1
[2]
[3 [6] [7]]
]
\end{forest}
\end{columns}
\end{frame}
\begin{frame}[fragile]
\frametitle{试找出非完全二叉树}
\begin{forest}
[{}, for tree={fill=black}]
\end{forest} \hspace{3cm}
\begin{forest}
[{}, for tree={fill=black} [{}] [{}]]
\end{forest}
\begin{forest}
[{}, for tree={fill=black}
[{} [{}] [{}]]
[{}]]
\end{forest}
\begin{forest}
[{}, for tree={fill=black}
[{} [{} [{}] [{}]] [{}]]
[{}]]
\end{forest}
\begin{forest}
[{}, for tree={fill=black}
[{} [{}] [{}]]
[{} [{}] [{}]]]
\end{forest}
\begin{forest}
[{}, for tree={fill=black}
[{} [{} [{} [{}] [{}]] [{}]] [{}]]
[{}]]
\end{forest}
\begin{forest}
[{}, for tree={fill=black}
[{} [{} [{}] [{}, no edge, draw=none, fill=none]] [{}]]
[{} [{}] [{}]]]
\end{forest}
\end{frame}
\begin{frame}[fragile]
\frametitle{二叉树的性质}
\begin{easylist}
& 性质4:具有$n$个结点的完全二叉树的深度为:
\[ \biggl\lfloor {log_2 n} \biggr\rfloor + 1 \]
\end{easylist}
\small
对于完全二叉树,设深度为$k$, 由$2^{k-1}-1 < n \leq 2^k - 1$可知,
$2^{k-1} \leq n <2^k$, 则 $k-1 \leq log_2n < k$
(参考性质2)
\begin{columns}[T]
\begin{column}{0.45\linewidth}
\scalebox{0.6}{
\begin{forest}
[1, for tree={minimum width=2.5em, minimum height=2.5em}
[2 [4 [8] [9]] [5 [10] [11]]]
[3 [6 [12] [13]] [7 [14] [15]]]
]
\end{forest}
}
满二叉树
\end{column}
\begin{column}{0.45\linewidth}
\scalebox{0.6}{
\begin{forest}
[1, for tree={minimum width=2.5em, minimum height=2.5em}
[2 [4 [8] [9]] [5 [10] [11]]]
[3 [6 [12] [{}, no edge, draw=none]] [7]]
]
\end{forest}
}
完全二叉树
\end{column}
\end{columns}
\end{frame}
\begin{frame}[fragile]
\frametitle{测试}
\begin{columns}[T]
\column{0.6\textwidth}
对于完全二叉树:
\begin{enumerate}
\item 若完全二叉树有叶子结点出现在第$k$层,它可能还有(~~~~~~)层的叶子结点;
\item 若某结点的右子树的最大层次为$L$,则其左子树的最大层次为(~~~~~~~~);
\item 若按如图所示的编号方式,试求出编号为$i$的节点的父节点和子节点的编号.
\end{enumerate}
\column{0.36\textwidth}
\scalebox{0.6}{
\begin{forest}
[1
[2 [4 [8] [9]] [5 [10] [11]]]
[3 [6 [12] [{}, no edge,draw=none]] [7]]
]\
\end{forest}
}
\end{columns}
\pause
\begin{enumerate}
\item $k-1$, $k+1$
\item $L$ or $L+1$
\end{enumerate}
\end{frame}
\begin{frame}[fragile]
\frametitle{二叉树的性质}
\begin{center}
\scalebox{0.6} {
\begin{forest}
[1, name=A,for tree={}
[2, name=B [4 [8] [9]] [5 [10] [11]]]
[3, name=C [6 [12] [{}, no edge,draw=none] ] [7] ]
]
\draw[draw, dotted, thick] (B.north) ++(-2cm, 0.2cm)-- ++(9cm, 0cm)
node[above, right, yshift=0.5cm]{1};
\draw[draw, dotted, thick] (B.south) ++(-2cm, -0.2cm)-- ++(9cm, 0cm)
node[above, right, yshift=0.5cm]{2};
\draw[draw, dotted, thick] (B.south) ++(-2cm, -1.6cm)-- ++(9cm, 0cm)
node[above, right, yshift=0.5cm]{3} node[below, right, yshift=-0.5cm] {4};
\end{forest}
}
\end{center}
\begin{itemize}
\item 性质5: 对具有$n$个结点的完全二叉树的结点按层次顺序编号,对任意结点$i$有:
\begin{itemize}
\item (有关结点$i$的双亲)若$i=1$,则为二叉树的根结点, 没有双亲;否则双亲结点的
编号为: $\biggl\lfloor{\dfrac{i}{2}}\biggr\rfloor$
\item (有关结点$i$的孩子) 若$n<2 \cdot i$, 则结点$i$无左孩子;否则左孩子编号
是$2 \cdot i$。
\item 若$n<2 \cdot i+1$,则结点i无右孩子;否则右孩子编号为$2 \cdot i+1$
\end{itemize}
\end{itemize}
\end{frame}
\begin{frame}[t,fragile]
\frametitle{二叉树的存储结构:顺序存储}
\begin{columns}
\begin{column}[T]{0.45\textwidth}
\scalebox{0.6} {
\begin{forest}
[1, name=A,for tree={color=white,fill=black}
[2, for tree={dotted,color=black,fill=white}, [4 [8] [9]] [5 [10] [11]]]
[3, name=C [6 [12] [{}, no edge,draw=none, fill=none] ] [7] ]
]
\end{forest}
}
\vspace{2cm}
\scalebox{0.6}{
\begin{tikzpicture}[ n/.style={minimum size=0.8cm}]
\draw node[] (n0) {};
\foreach \x [evaluate = \x as \xp using int(\x-1)] in {1, ..., 12}
\draw node[n, right=0 of n\xp](n\x) {\xp};
\foreach \x in {2,4,5,8,9,10,11}
\draw node[n, above=0 of n\x, draw] (d\x) { $\wedge$};
\foreach \x/\y in {1/1,3/3,6/6,7/7,12/12}
\draw node[n, above=0 of n\x, draw,white, fill=black](d\x) {\y};
\end{tikzpicture}
}
\end{column}
\begin{column}[T]{0.55\textwidth}
\begin{minted}[bgcolor=yellow!20, fontsize=\scriptsize]{c}
#define MAX_SIZE 100
typedef int SqBiTree[MAX_SIZE];
SqBiTree bt;
\end{minted}
\begin{minted}[bgcolor=red!5, fontsize=\scriptsize]{java}
class SqBiTree {
//static int MAX_SIZE = 100;
//int[] data = new int[MAX_SIZE];
List<Integer> data = new ArrayList<>();
}
\end{minted}
\end{column}
\end{columns}
\begin{itemize}
\item 把结点安排成一个恰当的序列(编号),存储在数组中
\item 便于“随机存取”
\end{itemize}
\end{frame}
\begin{frame}[fragile]
\frametitle{二叉树的存储结构:顺序存储}
\begin{itemize}
\item 适用于完全二叉树
\end{itemize}
\begin{columns}
\column{0.5\textwidth}
\scalebox{0.8} {
\begin{forest}
[$a$, for tree={color=white,fill=black}
[{}, draw=none, no edge, fill=none]
[$b$
[{}, draw=none, no edge, fill=none]
[$c$ [{}, draw=none, no edge, fill=none] [$d$] ]
]
]
\end{forest}
}
\column{0.4\textwidth}
\scalebox{0.8} {
\begin{forest}
[1, for tree={color=white,fill=black}
[2 [4] [5 [6] [7]]]
[3]
]
\end{forest}
}
\end{columns}
\pause
\scalebox{0.7}{
\begin{tikzpicture}[ n/.style={minimum size=0.8cm}]
\draw node[n] (n0) {0};
\foreach \x [evaluate = \x as \xp using int(\x-1)] in {1, ..., 14}
\draw node[n, right=0 of n\xp](n\x) {$\x$};
\foreach \x in {1,3,4,5,7,8, ..., 13}
\draw node[n, above=0 of n\x, draw](d\x) {$0$};
\foreach \x/\y in {0/a,2/b,6/c,14/d}
\draw node[n, above=0 of n\x, draw, fill=red!10](d\x) {$\y$};
\end{tikzpicture}
}
\pause
\hspace{4cm}
\scalebox{0.7}{
\begin{tikzpicture}[ n/.style={minimum size=0.8cm}]
\draw node[n] (n0) {0};
\foreach \x [evaluate = \x as \xp using int(\x-1)] in {1, ..., 10}
\draw node[n, right=0 of n\xp](n\x) {$\x$};
\foreach \x in {5, ..., 8}
\draw node[n, above=0 of n\x, draw](d\x) {$0$};
\foreach \x/\y in {0/1,1/2,2/3,3/4,4/5,9/6,10/7}
\draw node[n, above=0 of n\x, draw, fill=red!10](d\x) {$\y$};
\end{tikzpicture}
}
\end{frame}
\begin{frame}[fragile]
\frametitle{二叉树的链式存储:二叉链表}
\begin{columns}[T]
\column{0.4\textwidth}
\scalebox{0.7}{
\begin{forest}
[1
[2 [{}, no edge, draw=none] [4 [5] [6]]]
[3]]
\end{forest}
}
\column{0.4\textwidth}
\includegraphics[width=0.6\textwidth]{dot/tree-store-bilink.pdf}
\end{columns}
\small
\begin{columns}
\column{0.48\textwidth}
\begin{minted}[fontsize=\scriptsize,bgcolor=red!5]{c}
typedef struct BiTNode { // C Code
ElemType data;
struct BiTNode *lchild, *rchild;
} BiTNode, *BiTree;
\end{minted}
\column{0.48\textwidth}
\begin{minted}[fontsize=\scriptsize,bgcolor=yellow!10]{c}
class BiTNode<T> { //Java Code
T data;
BiTNode lchild, rchild;
}
\end{minted}
\end{columns}
思考:含$n$个结点的二叉链表中有多少个空指针?
\pause
指针域一共有$2*n$个,分支共有$N-1$, 每个分支占用一个指针域,所以空指针数量为: $2*n - (n-1) = n+1$
\end{frame}
\begin{frame}[fragile]
\frametitle{二叉树的链式存储:三叉链表}
\begin{itemize}
\item 二叉链表不便查找父节点, 可加一个指向双亲的指针
\end{itemize}
\begin{columns}[T]
\column{0.5\textwidth}
\begin{minted}[fontsize=\scriptsize,bgcolor=red!5]{python}
class BiTNode:
def __init__(self, lc, rc, parent):
self.lc = lc
self.rc = rc
self.parent = parent
\end{minted}
\begin{minted}[fontsize=\scriptsize,bgcolor=yellow!10]{c}
class BiTNode<T> { //Java Code
T data;
BiTNode lchild, rchild, parent;
}
\end{minted}
\column{0.5\textwidth}
\scalebox{0.6}{
\begin{forest}
[1
[2 [{}, no edge, draw=none] [4 [5] [6]]]
[3]]
\end{forest}
}
\includegraphics[width=0.75\textwidth]{dot/tree-store-trilink.pdf}
\scriptsize 思考: 有n个结点的三叉链表有多少个空指针?
\pause
对于二叉链表,指针域一共有$2*n$个,分支共有$N-1$, 每个分支占用一个指针域,所以空指针数量为$2*n - (n - 1) = n+1 $; 对于parent,根节点无父节点,所以共有$n+1+1=n+2$个空指针
\end{columns}
\end{frame}
\begin{frame}[fragile, plain]
~
\end{frame}
\subsection{遍历二叉树和线索二叉树}
\begin{frame}[fragile]
\frametitle{遍历二叉树和线索二叉树}
在二叉树的一些应用中,常常要求在树中查找具有某种特征的结点,或者对树中全部结点逐
一进行某种处理。这就引入了遍历二叉树的问题,即如何按某条搜索路径巡访树中的每一个
结点,使得每一个结点均被访问一次且仅访问一次。
\end{frame}
\begin{frame}[fragile]
\frametitle{二叉树的遍历}
\begin{itemize}
\item 遍历是指按某种方式访问所有结点,使每个结点被访问一次且只被访问一次。
\item 二叉树的遍历是按一定规则将二叉树的结点排成一个线性序列,即非线性序列线性
化。
\item 遍历的方式: 深度优先和广度优先,深度优先又分为三种:
\begin{itemize}
\item 先序次序
\item 中序次序(对称序次序)
\item 后序次序
\end{itemize}
\end{itemize}
\end{frame}
\begin{frame}[fragile]
\frametitle{二叉树的遍历}
\begin{easylist}
& 1、先序遍历二叉树的操作定义为: 若二叉树为空,则空操作;否则
&& (1)访问根结点; (2)先序遍历左子树; (3)先序遍历右子树。
& 2、中序遍历二叉树的操作定义为:若二叉树为空,则空操作;否则
&& (1)中序遍历左子树; (2)访问根结点; (3)中序遍历右子树。
& 3、后序遍历二叉树的操作定义为:若二叉树为空,则空操作;否则
&& (1)后序遍历左子树; (2)后序遍历右子树; (3)访问根结点。
\end{easylist}
\end{frame}
\begin{frame}[fragile]
\frametitle{示例}
\scalebox{0.8}{
\begin{forest}
[A
[B [D] [{}, no edge, draw=none]]
[C [E [{}, no edge, draw=none] [G]] [F [H] [I]]]
]
\end{forest}
}
\pause
\begin{itemize}
\item 先序:ABDCEGFHI
\item 中序:DBAEGCHFI
\item 后序:DBGEHIFCA
\item 广度优先:ABCDEFGHI
\end{itemize}
\end{frame}
\begin{frame}[fragile]
\frametitle{二叉树的遍历}
\begin{columns}[T]
\column{0.6\textwidth}
//先序遍历二叉树递归算法C伪代码
\begin{minted}[bgcolor=red!5, fontsize=\small]{c}
status preOrderTraverse(BiTree T){
if(T){
printf(T->data);
preOrderTraverse(T->lchild);
preOrderTraverse(T->rchild);
}
}
\end{minted}
\begin{itemize}
\item status代表什么?
\item printf代表什么?
\end{itemize}
\column{0.4\textwidth}
\scalebox{0.8}{
\begin{forest}
[A
[B [D] [{}, no edge, draw=none]]
[C [E [{}, no edge, draw=none] [G]] [F [H] [I]]]
]
\end{forest}
}
\end{columns}
\end{frame}
\begin{frame}[fragile]
\frametitle{二叉树的遍历}
\begin{columns}[T]
\column{0.6\textwidth}
先序遍历递归算法的Java实现
\begin{minted}[bgcolor=yellow!10,fontsize=\small]{java}
class Node {
String data;
Node left, right;
}
class Tree {
void preOrderTraverse(Node node) {
if(node != null) {
System.out.println(node.data);
preOrderTraverse(node.left);
preOrderTraverse(node.right);
}
}
}
\end{minted}
\column{0.4\textwidth}
\scalebox{0.8}{
\begin{forest}
[A
[B [D] [{}, no edge, draw=none]]
[C [E [{}, no edge, draw=none] [G]] [F [H] [I]]]
]
\end{forest}
}
\end{columns}
\end{frame}
\begin{frame}[fragile, allowframebreaks]
\frametitle{中序遍历的Java实现}
\begin{minted}[fontsize=\scriptsize]{java}
public class Tree {
Node root;
public Tree(Node root) {
this.root = root;
}
void inOrderTraverse() {
inOrderTraverse(root);
}
void inOrderTraverse(Node node) {
if(node != null) {
inOrderTraverse(node.lc);
//visit(node);
System.out.println(node);
inOrderTraverse(node.rc);
}
}
public static void main(String[] args) {
Node a =new Node("A",
new Node("B", new Node("D"), null),
new Node("C", new Node("E"), new Node("F"))
);
Tree tree = new Tree(a);
tree.inOrderTraverse();
}
static class Node {
String data;
Node lc, rc;
public Node(String data, Node lc, Node rc) {
this.data = data;
this.lc = lc;
this.rc = rc;
}
public Node(String data) {
this.data = data;
this.lc = null;
this.rc = null;
}
public String toString(){
return data;
}
}
}
\end{minted}
\end{frame}
\begin{frame}[fragile]
\frametitle{如下将得到树的什么序列?}
\begin{columns}[T]
\column{0.6\textwidth}
\begin{minted}[bgcolor=yellow,fontsize=\small]{c}
status traverse(BiTree T) {
InitStack(S);
p = T;
while(p || !StackEmpty(S)) {
if(p) {
push(S, p); p = p->lchild;
} else {
pop(S, p);
printf(p->data);
p = p->rchild;
}
}
return OK;
}
\end{minted}
\column{0.4\textwidth}
\scalebox{0.8}{
\begin{forest}
[A
[B [D] [{}, no edge, draw=none]]
[C [E [{}, no edge, draw=none] [G]] [F [H] [I]]]
]
\end{forest}
}
\end{columns}
\end{frame}
\begin{frame}[fragile]
\frametitle{如下将得到树的什么序列?}
\begin{columns}[T]
\column{0.6\textwidth}
\begin{minted}[bgcolor=yellow!10,fontsize=\small]{c}
void traverse(TreeNode root) {
Stack stack = new Stack();
TreeNode p = root;
while(p!=null || stack.notEmpty()) {
if(p!=null) {
stack.push(p);
p = p.lchild;
} else {
p = stack.pop();
printf(p.data);
p = p.rchild;
}
}
}
\end{minted}
\column{0.4\textwidth}
\scalebox{0.8}{
\begin{forest}
[A
[B [D] [{}, no edge, draw=none]]
[C [E [{}, no edge, draw=none] [G]] [F [H] [I]]]
]
\end{forest}
}
\end{columns}
\end{frame}
\begin{frame}[fragile]
\frametitle{二叉树的遍历:广度优先}
\begin{columns}[T]
\column{0.6\textwidth}
算法步骤:
\begin{itemize}
\item 访问节点
\item 从左到右依次访问儿子节点
\item 重复前一步骤
\end{itemize}
\begin{tikzpicture}[ n/.style={minimum width=1cm,minimum height=0.6cm, draw}]
\draw node[] (n0) {};
\foreach \x [evaluate = \x as \xp using int(\x-1)] in {1, ..., 7}
\draw node[n, right=0 of n\xp](n\x) {};
\draw node[n, right=0 of n0] {A};
\end{tikzpicture}
\begin{tikzpicture}[ n/.style={minimum width=1cm,minimum height=0.6cm, draw}]
\draw node[] (n0) {};
\foreach \x [evaluate = \x as \xp using int(\x-1)] in {1, ..., 7}
\draw node[n, right=0 of n\xp](n\x) {};
\draw node[n, right=0 of n1] {B} node[n, right=0 of n2]{C};
\end{tikzpicture}
\begin{tikzpicture}[ n/.style={minimum width=1cm,minimum height=0.6cm, draw}]
\draw node[] (n0) {};
\foreach \x [evaluate = \x as \xp using int(\x-1)] in {1, ..., 7}
\draw node[n, right=0 of n\xp](n\x) {};
\draw node[n, right=0 of n2] {C} node[n, right=0 of n3]{D};
\end{tikzpicture}
.......
课堂练习:编程实现广度优先遍历。
\column{0.4\textwidth}
\scalebox{0.8}{