MIT算法导论课程：Lec09 二叉搜索树，对应书上的章节：Section 12.4
1. Binary-search-tree sort
二叉搜索树排序过程如下：
遍历树的时间为\(O(n)\)，但是构建BST的时间是多少？
BST sort和quick sort很像：BST sort performs the same comparisons as quicksort, but in a different order!
2. Expected node depth
Assuming all input permutations are equally likely, we have：（BST sort performs the same comparisons as quicksort） \[ Average\ node\ depth=\frac{1}{n}E[\sum(comparisons\ to\ insert\ node\ i)]=\frac{1}{n}O(n\lg n)=O(\lg n) \] 但是平均节点深度并不等于树的期望高度。
3. Analyzing height
Jensen’s inequality： \(f(E[X])\leq E[f(x)]\) for any convex function f and random variable X.
Exponential height： random variable \(Y_n=x^{X_n}\)，where \(X_n\) is the random varible denoting the height of BST.
证明： Prove that \(2^{E[X_n]}\leq E[2^{X_n}]=E[Y_n]=O(n^3)\),hence that \(E[X_n]=O(\lg n)\)
凸函数(Convex functions)定义： \(f:R->R\) is convex if for all \(\alpha,\beta\geq 0\) such that \(\alpha+\beta=1\),we have \(f(\alpha x+\beta y)\leq \alpha f(x)+\beta f(y)\) for all \(x,y\in R\).
Convexity lemma： Let \(f\) be a convec function,and let \(\alpha_1,..,\alpha_n\) be nonnegative real numbers such that\(\sum \alpha_k=1\), Then for any real numbers \(x_1,...,x_n\),we have\(f\left(\sum_{k=1}^{n} \alpha_{k} x_{k}\right) \leq \sum_{k=1}^{n} \alpha_{k} f\left(x_{k}\right)\).
proof: 归纳法证明： \[ \begin{aligned} f\left(\sum_{k=1}^{n} \alpha_{k} x_{k}\right) &=f\left(\alpha_{n} x_{n}+\left(1-\alpha_{n}\right) \sum_{k=1}^{n-1} \frac{\alpha_{k}}{1-\alpha_{n}} x_{k}\right) \\ & \leq \alpha_{n} f\left(x_{n}\right)+\left(1-\alpha_{n}\right) f\left(\sum_{k=1}^{n-1} \frac{\alpha_{k}}{1-\alpha_{n}} x_{k}\right) \\ & \leq \alpha_{n} f\left(x_{n}\right)+\left(1-\alpha_{n}\right) \sum_{k=1}^{n-1} \frac{\alpha_{k}}{1-\alpha_{n}} f\left(x_{k}\right)\\ &=\sum_{k=1}^{n} \alpha_{k} f\left(x_{k}\right)\\ \end{aligned} \]
Convexity lemma: infinite case
引申到 Jensen’s inequality：
Analysis of BST height
Let Xn be the random variable denoting the height of a randomly built binary search tree on n nodes, and let \(Y_n = 2^{X_n}\) be its exponential height.
If the root of the tree has rank k, then \(X_n=1+max\{X_{k-1},X_{n-k}\}\).Hence, we have\(Y_n=2*\{maxY_{k-1},Y_{n-k}\}\).
Define the indicator random variable \(Z_{nk}\) as :\(Z_{n k}=\left\{\begin{array}{ll} 1 & \text { if the root has rank } k \\ 0 & \text { otherwise. } \end{array}\right.\)
Thus, \(\operatorname{Pr}\{Z_{n k}=1\}=\mathrm{E}[Z_{n k}]=1/ n\), and \(Y_{n}=\sum_{k=1}^{n} Z_{n k}(2 \cdot \max \{Y_{k-1}, Y_{n-k}\})\).
\[ \begin{aligned} E[Y_{n}] &=E[\sum_{k=1}^{n} Z_{n k}(2 \cdot \max \{Y_{k-1}, Y_{n-k}\})] \\ &=\sum_{k=1}^{n} E[Z_{n k}(2 \cdot \max \{Y_{k-1}, Y_{n-k}\})] \\ &=2 \sum_{k=1}^{n} E[Z_{n k}] \cdot E[\max \{Y_{k-1}, Y_{n-k}\}] \\ & \leq \frac{2}{n} \sum_{k=1}^{n} E[Y_{k-1}+Y_{n-k}] \\ &=\frac{4}{n} \sum_{k=0}^{n-1} E\left[Y_{k}\right] & \begin{array}{c} \text { Each term appears } \\ \text { twice, and reindex } \end{array} \end{aligned} \]
采用替代法证明\(E[Y_n]\leq cn^3\) for some positive constant.
\[ \begin{aligned} E\left[Y_{n}\right] &=\frac{4}{n} \sum_{k=0}^{n-1} E\left[Y_{k}\right] \\ & \leq \frac{4}{n} \sum_{k=0}^{n-1} c k^{3} \\ & \leq \frac{4 c}{n} \int_{0}^{n} x^{3} d x \\ &=\frac{4 c}{n}\left(\frac{n^{4}}{4}\right)\\ &=\mathrm{cn}^{3} \end{aligned} \]
Putting it all together, we have\(2^{E\left[X_{n}\right]} \leq E\left[2^{X_{n}}\right]=E[Y_n]\leq cn^3\),since \(f(x)=2^x\)is convex
可得\(E[X_n]\leq 3\lg n+O(1)\).
4. Post mortem
