leetCode 235. Lowest Common Ancestor of a Binary Search Tree 二排序树问题

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：

mine思路：

1.将各个节点的从root到节点的路径都记录下来vector<vector<int>>，vector中最后一个元素为当前节点。

2.找到目标的两个vector，比较之找到较短的"根"

others思路：

在二叉查找树种，寻找两个节点的最低公共祖先。

1、如果a、b都比根节点小，则在左子树中递归查找公共节点。

2、如果a、b都比根节点大，则在右子树中查找公共祖先节点。

3、如果a、b一个比根节点大，一个比根节点小，或者有一个等于根节点，则根节点即为最低公共祖先。

代码如下：(代码实现为others思路)

/**
 * 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) {
        int pVal = p->val;
        int qVal = q->val;
        int rootVal = root->val;
        if(pVal < rootVal && qVal < rootVal)
        {
            return lowestCommonAncestor(root->left,p,q);
        }
        else if(pVal > rootVal && qVal > rootVal)
        {
            return lowestCommonAncestor(root->right,p,q);
        }
        return root;
    }
};

2016-08-07 09:47:25