236. Lowest Common Ancestor of a Binary Tree

最新推荐文章于 2024-02-06 06:38:28 发布

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本文介绍三种寻找二叉树中两个节点的最近公共祖先的方法：通过建立包含指向父节点的树来查找、使用分治法检查节点是否分别位于两棵子树中、直接找到节点路径并确定公共祖先。

方法1:

建立指向父节点的树；

struct Node{
    Node* pre;
    Node* left;
    Node* right;
    TreeNode* cur;
    
    Node(TreeNode* root,Node* pre_root):cur(root),pre(pre_root){}
};
class Solution {
public:
    //建立包含指向父节点的树
    void DFS(TreeNode* root,Node* pre_root)
    {
        if(root==NULL)
            return;
        
        Node* pre_left=new Node(root->left,pre_root);
        Node* pre_right=new Node(root->right,pre_root);
        pre_root->left=pre_left;
        pre_root->right=pre_right;
        DFS(root->left,pre_left);
        DFS(root->right,pre_right);
    }
    
    //目标节点在父节点树中的位置
    void DFS_Node(Node* root,Node *&root_p,TreeNode* p)
    {
        if(root==NULL||root->cur==NULL)
            return;
        if(root->cur==p)
        {
            root_p=root;
            return;
        }
        DFS_Node(root->left,root_p,p);
        DFS_Node(root->right,root_p,p);
    }
    
    //根据父节点树找到从根节点到目标节点的路径
    stack<TreeNode*> GetRoute(Node* root)
    {
        stack<TreeNode*> vec;
        while(root->pre!=NULL)
        {
            vec.push(root->cur);
            root=root->pre;
        }
        vec.push(root->cur);
        return vec;
    }
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(!root)
            return NULL;
        Node* pre_root=new Node(root,NULL);
        DFS(root,pre_root);
        Node* pre_p, *pre_q;
        DFS_Node(pre_root, pre_p, p);
        DFS_Node(pre_root, pre_q, q);
        
        stack<TreeNode*> res_p,res_q;
        res_p=GetRoute(pre_p);
        res_q=GetRoute(pre_q);
        
        
        TreeNode* top=NULL;
        while(!res_p.empty()&&!res_q.empty())
        {
            if(res_p.top()!=res_q.top())
                break;
            top=res_p.top();
            res_p.pop();
            res_q.pop();
        }
        return top;
    }
};

方法2:

分治法，查看节点是否在两颗子树中；

class Solution {
public:

    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(root==NULL)
            return NULL;
        if(root==p||root==q)return root;
        TreeNode* left=lowestCommonAncestor(root->left, p, q);
        TreeNode* right=lowestCommonAncestor(root->right, p, q);
        if(left&&right)
            return root;
        return left==NULL?right:left;
    }
};

方法3:

直接找到节点路径

class Solution {
public:
    
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q)
    {
        if(root==NULL||q==NULL||p==NULL)
            return NULL;
        
        vector<TreeNode*> vp,vq,res_p,res_q;
        vp.push_back(root);
        vq.push_back(root);
        
        GetPath(root, p,vp);
        GetPath(root, q,vq);
        
        
        TreeNode* lca=NULL;
        
        for(int i=0;i<vp.size()&&i<vq.size();i++)
        {
            if(vp[i]==vq[i])lca=vp[i];
            else
                break;
        }
        return lca;
    }
private:
    bool GetPath(TreeNode* root,TreeNode* p, vector<TreeNode*>& vt)
    {
        if(root==p)
        {
            return true;
        }
        if(root->left)
        {
            vt.push_back(root->left);
            if(GetPath(root->left, p, vt)) return true;
            vt.pop_back();
        }
        if(root->right)
        {
            vt.push_back(root->right);
            if(GetPath(root->right, p, vt)) return true;
            vt.pop_back();
        }
        return false;
    }
};

其他：

http://arsenal591.blog.163.com/blog/static/253901269201510169448656/