Lowest Common Ancestor of a Binary Search Tree-优快云博客

本文链接：https://blog.youkuaiyun.com/u010786672/article/details/46844747

本文介绍了一种方法，在给定的二叉搜索树中找到两个指定节点的最近公共祖先。通过实现一个类和相关函数，我们能够高效地解决这一问题，遵循了二叉搜索树的性质来优化查找过程。

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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.

/**
 * 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){
            return NULL;
        }
        vector<TreeNode*> v1;
        vector<TreeNode*> v2;
        find(root,p,v1);//找p的前缀路径
        find(root,q,v2);//找q的前缀路径
        int len=min(v1.size(),v2.size());
        TreeNode* last=NULL;
        for(int i=0;i<len;i++){
            if(v1[i]==v2[i]){
                last=v1[i];
            }else{
                break;
            }
        }
        return last;
    }

    void find(TreeNode* root, TreeNode* p, vector<TreeNode*> &v){
        if(root==NULL){
            return;
        }
        if(root==p){
            v.push_back(root);
            return;
        }
        v.push_back(root);
        find(root->left,p,v);
        if(v[v.size()-1]==p)//左子树中,找到了p的前缀路径
            return;
        v.pop_back();

        v.push_back(root);
        find(root->right,p,v);
        if(v[v.size()-1]==p)//右子树中,找到了p的前缀路径
            return;
        v.pop_back();

    }
};