235. Lowest Common Ancestor of a Binary Search Tree


1. 問題描述

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.

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Tags: Tree
Similar Problems: (M) Lowest Common Ancestor of a Binary Tree

2. 解題思路

  • 首先,提取從根節點到指定節點的路徑;
  • 然后,轉換成求兩個鏈表中第一個公共節點的問題!

3. 代碼

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q)
    {
        if (root == NULL || p == NULL || q == NULL)
        {
            return NULL;
        }

        std::stack<TreeNode *> staPath_p;
        std::stack<TreeNode *> staPath_q;
        GetBinaryTreePath(root, p, staPath_p);
        GetBinaryTreePath(root, q, staPath_q);

        int pSize = staPath_p.size();
        int qSize = staPath_q.size();

        if (pSize > qSize)
        {
            for (int i=0; i<pSize-qSize; i++)
            {
                staPath_p.pop();
            }        
        }
        else
        {
            for (int i=0; i<qSize-pSize; i++)
            {
                staPath_q.pop();
            } 
        }

        while (staPath_p.top() != staPath_q.top())
        {
            staPath_p.pop();
            staPath_q.pop();
        }
        return staPath_p.top();
    }

private:
    bool GetBinaryTreePath(TreeNode* root, TreeNode* pDesNode, std::stack<TreeNode *> &staPath)
    {
        bool bIsFind = false;
        staPath.push(root);
        if (root == pDesNode)
        {            
            return true;
        }

        if (NULL == staPath.top()->left && NULL == staPath.top()->right)
        {
            staPath.pop();
            return false;
        }

        if (NULL != root->left)
        {
            bIsFind = GetBinaryTreePath(root->left, pDesNode, staPath);
        }
        if ( NULL != root->right && !bIsFind)
        {
            bIsFind = GetBinaryTreePath(root->right, pDesNode, staPath);
        }

        if (false == bIsFind)
        {
            staPath.pop();
            return false;
        }
        else
        {
            return true;
        }
    }
};

4. 反思


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