The-Art-Of-Programming-By-July/03.03.md at master ・ julycoding/The-Art-Of-Programming-By-July ・ GitHub



The-Art-Of-Programming-By-July/03.03.md at master ・ julycoding/The-Art-Of-Programming-By-July ・ GitHub

最近公共祖先LCA问题 问题描述 求有根树的任意两个节点的最近公共祖先。 分析与解法 举个例子,如针对下图所示的一棵普通的二叉树来讲: 结点3和结点4的最近公共祖先是结点2,即LCA(3 4)=2 。在此,需要注意到当两个结点在同一棵子树上的情况,如结点3和结点2的最近公共祖先为2,即 LCA(3,2)=2。同理:LCA(5,6)=4,LCA(6,10)=1。 明确了题意,咱们便来试着解决这个问题。直观的做法,可能是针对是否为二叉查找树分情况讨论,这也是一般人最先想到的思路。除此之外,还有所谓的Tarjan算法、倍增算法、以及转换为RMQ问题(求某段区间的极值)。后面这几种算法相对高级,不那么直观,但思路比较有启发性,了解一下也有裨益。 解法一:暴力对待 1.1、是二叉查找树 在当这棵树是二叉查找树的情况下,如下图: 如果当前结点t 大于结点u、v,说明u、v都在t 的左侧,所以它们的共同祖先必定在t 的左子树中,故从t 的左子树中继续查找; 如果当前结点t 小于结点u、v,说明u、v都在t 的右侧,所以它们的共同祖先必定在t 的右子树中,故从t 的右子树中继续查找; 如果当前结点t 满足 u right) { int temp = left; left = right; right = temp; } while (true) { //如果t小于u、v,往t的右子树中查找 if (t.value < left) { parent = t; t = t.right; //如果t大于u、v,往t的左子树中查找 } else if (t.value > right) { parent = t; t = t.left;

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