[leetcode] 235. Lowest Common Ancestor of a Binary Search Tree

Question :

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.

Solution :

因为是BST，因此要找LCA就简单很多，从结果反过来看，如果一个节点是p和q的LCA，那么这个节点就会比p大或相等且比q小或相等，或者反过来，而如果pq都比当前节点大，则说明满足前一句话的节点会在当前节点的右子树中，否则就在左子树中。
总而言之，判断p和q是否位于root的两端，是则为LCA，否则通过BST的特性确定p和q在左子树或右子树，然后递归找到LCA。

/** * 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) || (p->val <= root->val && q->val >= root->val))            return root;        if (p->val >= root->val && q->val >= root->val)            return lowestCommonAncestor(root->right, p, q);        return lowestCommonAncestor(root->left, p, q);    }};