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leetcode:Lowest Common Ancestor of a Binary Search Tree

时间:2015-07-17 11:20:19      阅读:115      评论:0      收藏:0      [点我收藏+]

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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.

分析:由于在所谓的二叉搜索树(binary search tree)中,处处满足顺序性(即任一节点的左(右)子树种,所有节点均小于(大于)r)。

思路: 1、当头结点为空时,返回空指针

          2、如果节点p、q的值都比root的值要小,那么LCA一定为root的左子树;而如果节点p、q的值都比root的值要大,那么LCA一定为root的右子树;当节点p、q的值中一个比root的值要大,另一个比它小时,LCA就是root了。

          3、我们要考虑是否能覆盖到节点是它自身子节点的情况,这时返回的是p或q的其中一个。

代码如下:(recursive solution)

/**
 * 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) {
        while(root !=nullptr){
        if(p->val < root->val && q->val < root->val){
        return lowestCommonAncestor(root->left,p,q);
        }
        else if(p->val > root->val && q->val > root->val){
        return lowestCommonAncestor(root->right,p,q);
        }
        else return root;     // 当p->val = root->val或 q->val =root->val时,就是节点是它自身子节点的情况了
}
return root;
}
};

 也可以简洁点:

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;
    }
};

  

其他参考解法:

class Solution {
public:
     TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(p->val > root->val && q->val < root->val)
        {
            return root;
        }
        if(p->val < root->val && q->val > root->val)
        {
            return root;
        }
        if( p->val == root->val || q->val == root->val)
            return root;

        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);
    }
};

或:(iterative solution)  

class Solution { 
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        TreeNode* cur = root;
        while (true) {
            if (p -> val < cur -> val && q -> val < cur -> val)
                cur = cur -> left;
            else if (p -> val > cur -> val && q -> val > cur -> val)
                cur = cur -> right;
            else return cur; 
        }
    }
};

  

 

 

leetcode:Lowest Common Ancestor of a Binary Search Tree

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原文地址:http://www.cnblogs.com/carsonzhu/p/4653793.html

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