用順序存儲結構存儲n個結點的完全二叉樹編號為i的結點其雙親編號是ëi/û(i=時無雙親)其左子女是i(若i<=n否則i無左子女)右子女是i+(若i+<=n否則無右子女)
根據完全二叉樹的性質最後一個結點（編號為n）的雙親結點的編號是ën/û這是最後一個分枝結點在它之後是第一個終端（葉子）結點故序號最小的葉子結點的下標是ën/û+
按前序遍歷對頂點編號即根結點從１開始對前序遍歷序列的結點從小到大編號
設樹的結點數為n分枝數為B則下面二式成立
n=n+n+n+…+nm ()
n=B+= n+n+…+mnm ()
由()和()得葉子結點數n=+
élognù +
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From:http://tw.wingwit.com/Article/program/sjjg/201311/22665.html
