236. Lowest Common Ancestor of a Binary Tree

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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/

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