1.题目描述

2.算法思路

2.1算法思路1

2.2算法思路2

思想很简单，但是最难的是怎么用栈来记录q、p的路线。所以下面才是关键。

3.代码实现

3.1代码实现1

class Solution {
public:
    bool find(TreeNode* root,TreeNode* obj)
    {
        if(root==nullptr)
        {
            return false;
        }
        if(root->val==obj->val)
        {
            return true;
        }
        return find(root->left,obj)||find(root->right,obj);
    }
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) 
    {
        if(root==p||root==q)
        {
            return root;
        }
        bool pinright,pinleft,qinright,qinleft;
        pinleft=find(root->left,p);
        pinright=!pinleft;
        qinleft=find(root->left,q);
        qinright=!qinleft;

        if((qinleft&&pinright)||(qinright&&pinleft))
        {
            return root;
        }

        if(qinleft&&pinleft)
        {
            return lowestCommonAncestor(root->left,p,q);
        }   

        if(pinright&&qinright)
        {
            return lowestCommonAncestor(root->right,p,q);

        }
        return nullptr;
    }
};

3.2 代码实现2

class Solution {
public:
    bool find(TreeNode* root,TreeNode* cur,stack<TreeNode*>& path)
    {
        //根为空
        if(root==nullptr)
        {
            return false;
        }
        //根入栈
        path.push(root);
        //根即为所求
        if(root==cur)
        {
            return true;
        }

        //递归去左子树  如果左子树找到了就返回true
        if(find(root->left,cur,path))
        {
            return true;
        }

        //递归去右子树  如果右子树找到就返回true
        if(find(root->right,cur,path))
        {
            return true;
        }

        //左子树右子树都没找到 pop根  返回false
        path.pop();
        return false;



    }
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) 
    {
        stack<TreeNode*> ppath;
        stack<TreeNode*> qpath;

        find(root,p,ppath);
        find(root,q,qpath);

        while(ppath.size()>qpath.size())
        {
            ppath.pop();
        }

        while(qpath.size()>ppath.size())
        {
            qpath.pop();
        }

        while(ppath.size())
        {
            if(ppath.top()==qpath.top())
            {
                return ppath.top();
            }
            ppath.pop();
            qpath.pop();
        }
        return nullptr;
    }
};