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  • Leetcode题目:Lowest Common Ancestor of a Binary Search Tree

    题目: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.

    题目解答:这个题目非常简单,就是求在二叉搜索树中,两个子节点的最近公共祖先。

    首先说明一下,在二叉排序树中,是没有取值相同的元素的。另外,它还具有以下特点:

    二叉排序树或者是一棵空树,或者是具有下列性质的二叉树:
    (1)若左子树不空,则左子树上所有结点的值均小于它的根结点的值;
    (2)若右子树不空,则右子树上所有结点的值均大于它的根结点的值;
    (3)左、右子树也分别为二叉排序树;

    代码如下:

    /**
     * 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((root == NULL) || (p == NULL) || (q == NULL))
                return NULL;
            TreeNode * curNode = root;
            if(p -> val > q -> val)
            {
                while(curNode != NULL)
                {
                    if(curNode -> val > p -> val)
                    {
                        curNode = curNode -> left;
                    }
                    else if(curNode -> val < q -> val)
                    {
                        curNode =curNode -> right;
                    }
                    else
                        return curNode;
                }
            }
            else if(p -> val < q -> val)
            {
                while(curNode != NULL)
                {
                    if(curNode -> val > q -> val)
                    {
                        curNode = curNode -> left;
                    }
                    else if(curNode -> val < p -> val)
                    {
                        curNode =curNode -> right;
                    }
                    else
                        return curNode;
                }
            }
            return NULL;
        }
    };

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  • 原文地址:https://www.cnblogs.com/CodingGirl121/p/5414225.html
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