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Lowest Common Ancestor of a Binary Search Tree
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.
经典问题！先得到从根节点到两个点的两条路径，下面的问题就是求两个链表相交的位置了。不过因为元素都不相同，可以用队列存路径，从根节点向下走，每一个分叉的地方就是所求的LCA！
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 void getPath(TreeNode *root, TreeNode *p, queue<TreeNode*> &path) { 13 TreeNode *tmp = root; 14 while (tmp != p) { 15 path.push(tmp); 16 if (tmp->val > p->val) tmp = tmp->left; 17 else tmp = tmp->right; 18 } 19 path.push(p); 20 } 21 TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { 22 queue<TreeNode*> path1, path2; 23 getPath(root, p, path1); 24 getPath(root, q, path2); 25 TreeNode *res = root; 26 while (!path1.empty() && !path2.empty()) { 27 if (path1.front() == path2.front()) { 28 res = path1.front(); 29 path1.pop(); 30 path2.pop(); 31 } else { 32 break; 33 } 34 } 35 return res; 36 } 37 };
[LeetCode] Lowest Common Ancestor of a Binary Search Tree
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原文地址：http://www.cnblogs.com/easonliu/p/4639043.html
