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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.
这个题目的意思就是：让你找出两个结点最近的祖先，也就是和他们血缘关系最近的祖先
我的解法：深度优先搜索，再以这个结点，进行搜索，如果以这个结点进行搜索时，发现了我们要求的两个结点，则该结点就是这两个结点最近的 ，不过我的算法的时间复杂度很高。。。。
/** * 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: void dfs(TreeNode *root, TreeNode *p, TreeNode *q,int &i) { if (root == NULL)return; if (root == p || root == q) ++i; dfs(root->left, p, q,i); dfs(root->right, p, q,i); } TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { map<TreeNode*,int> visited; stack<TreeNode*> sttn; TreeNode * curNode; sttn.push(root); while (!sttn.empty()) { curNode = sttn.top(); while(curNode->left != NULL&&visited[curNode->left] != 1) { curNode = curNode->left; sttn.push(curNode); } if (curNode->right != NULL&&visited[curNode->right] != 1) { curNode = curNode->right; sttn.push(curNode); } else { int cnt = 0; dfs(curNode, p, q,cnt); if (cnt == 2) return curNode; visited[curNode] = 1; sttn.pop(); } } } };
LeetCode Lowest Common Ancestor of a Binary Serach Tree
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原文地址：http://www.cnblogs.com/csudanli/p/5327606.html
