[leetcode-235-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.

思路：

二叉查找树，左子树的值比根值小，右子树的值比根值大。如果p q都比root小，说明在左子树上，如果pq的值都比root大说明在右子树上，如果一左一右，则root即为所求。

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

 

参考：

https://discuss.leetcode.com/topic/18614/easy-c-recursive-and-iterative-solutions