［抄题］：
Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
“The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”
_______3______
/ ___5__ ___1__
/ \ / 6 _2 0 8
/ 7 4
For example, the lowest common ancestor (LCA) of nodes 5 and 1 is 3. Another example is LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.
［思维问题］：
不知道子节点怎么用dc。直接对给出的p,q节点进行操作即可。
［一句话思路］：
对pq节点做分合法定义。
［输入量］：空： 正常情况：特大：特小：程序里处理到的特殊情况：异常情况（不合法不合理的输入）：
［画图］：
［一刷］：
(left != null && right != null) 时，返回的是root节点的结果，不需要再做递归运算了。是一个“合”的过程。
［二刷］：
［三刷］：
［四刷］：
［五刷］：
［总结］：
［复杂度］：Time complexity: O(n) Space complexity: O(n)
［英文数据结构，为什么不用别的数据结构］：
只有dc算法，没有数据结构
［其他解法］：
自己写traverse函数：不好，会形成全局变量
［Follow Up］：
［LC给出的题目变变变］：
Lowest Common Ancestor of a Binary Search Tree 一模一样的，约束条件没用，直接套。
public class Solution { /* * @param root: The root of the binary search tree. * @param A: A TreeNode in a Binary. * @param B: A TreeNode in a Binary. * @return: Return the least common ancestor(LCA) of the two nodes. */ public TreeNode lowestCommonAncestor(TreeNode root, TreeNode A, TreeNode B) { if (root == null || A == root || B == root) {// return root; } //divide TreeNode left = lowestCommonAncestor(root.left, A, B); TreeNode right = lowestCommonAncestor(root.right, A, B); //conquer if (left != null && right != null) { return root;// } else if (left != null) { return left; } else if (right != null) { return right; } else { return null; } } }
