235. Lowest Common Ancestor of a Binary Search Tree
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题目:
Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.
According to the definition of LCA on Wikipedia: “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).”
_______6______ / \ ___2__ ___8__ / \ / \ 0 _4 7 9 / \ 3 5
For example, the lowest common ancestor (LCA) of nodes 2
and 8
is 6
. Another example is LCA of nodes 2
and 4
is 2
, since a node can be a descendant of itself according to the LCA definition.
本题题目出得花里胡哨的,其实这类BTS的题目都是毫无意义的,只要抓住递归的本质,首先做这类与数据结构密切相关的题目时,一定要熟悉BTS的性质:
(1)若左子树不空,则左子树上所有结点的值均小于或等于它的根结点的值;
(2)若右子树不空,则右子树上所有结点的值均大于或等于它的根结点的值;
(3)左、右子树也分别为二叉排序树;
代码:/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public: TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { if(p->val<root->val&&q->val<root->val) return lowestCommonAncestor(root->left,p,q); else if(p->val>root->val&&q->val>root->val) return lowestCommonAncestor(root->right,p,q); else return root; }};
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