# Lowest Common Ancestor of a Binary Tree

Source: Internet
Author: User

Solution one: For a case that is too deep.

`/** * 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==NU        LL) return NULL;        P1=p;        p2=q;        DFS (root,1);            while (num1!=num2) {if (num1>num2) num1=num1>>1;        else num2=num2>>1;        } ansnum=num1;        GetNode (root,1);    return ansnode;    }private:int Num1,num2,ansnum;    TreeNode *p1,*p2,*ansnode;            void Dfs (TreeNode *root,int num) {if (root) {if (ROOT==P1) num1=num;            if (ROOT==P2) num2=num;            DFS (Root->left,2*num);        DFS (ROOT-&GT;RIGHT,2*NUM+1);            }} void GetNode (TreeNode *root,int num) {if (root) {if (num==ansnum) ansnode=root;            GetNode (Root->left,2*num); GetNode (root->RIGHT,2*NUM+1); }    }};`

Solution Two: All can be over.

`/** * 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) {        ans=null;< C6/>dfs (root,p,q);        return ans;    } Private:    TreeNode *ans;    treenode* dfs (TreeNode *root,treenode *p,treenode *q) {        if (root&&ans==null) {            treenode* P1=dfs ( ROOT->LEFT,P,Q);            treenode* P2=dfs (root->right,p,q);            BOOL X=false,y=false;            if (root==p| | p1==p| | p2==p) x=true;            if (root==q| | p1==q| | P2==Q) y=true;            if (x&&y) {                ans=root;                return NULL;            }            if (x) return p;            if (y) return q;            return NULL;        }        return NULL;    }};`

Lowest Common Ancestor of a Binary Tree

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