leetcode 236. Lowest Common Ancestor of a Binary Tree

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

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).”

        _______3______       /              \    ___5__          ___1__   /      \        /      \   6      _2       0       8         /  \         7   4

For example, the lowest common ancestor (LCA) of nodes 5 and 1 is 3. Another example is LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

第一种方法

struct TreeNode {      int val;      TreeNode *left;      TreeNode *right;      TreeNode(int x) : val(x), left(NULL), right(NULL) {}  };    class Solution {            void findpath(TreeNode* node, TreeNode* p,vector<TreeNode*>pathpp,vector<TreeNode*>&pathp)      {          if(node==p)          {              vector<TreeNode*>path=pathpp;              path.push_back(node);              pathp=path;              return ;          }          if(node->left!=NULL)          {              vector<TreeNode*>path=pathpp;              path.push_back(node);              if(node->left==p)              {                  path.push_back(p);                  pathp=path;                  return ;              }              else                  findpath(node->left,p,path,pathp);          }          if(node->right!=NULL)          {              vector<TreeNode*>path=pathpp;              path.push_back(node);              if(node->right==p)                  {                      path.push_back(p);                      pathp=path;                  return ;              }              else                  findpath(node->right,p,path,pathp);          }      }  public:      TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {          vector<TreeNode*>pathp,pp,qq,pathq;          findpath(root,p,pp,pathp);          findpath(root,q,qq,pathq);          int k=pathp.size()<pathq.size()?pathp.size():pathq.size();          int i=0;          while(i<k&&pathp[i]==pathq[i])              i++;          if(i==k)              return pathp.size()<pathq.size()?pathp.back():pathq.back();          else              return pathp[i-1];      }  };  

Memory Limit Exceeded

构造节点TN记录每个节点的父节点，否则向上找父节点太麻烦，val其实也可以不要。

struct TN {int val;TN*parent;TN(int x) : val(x),parent(NULL) {}};class Solution {public:TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {vector<TreeNode*>que, qq;que.push_back(root);TN*root1 = new TN(root->val);vector<TN*>que1, qq1; que1.push_back(root1);bool pf(true), qf(true);while (!que.empty()&&(pf||qf)){vector<TreeNode*>newque;vector<TN*>newque1;for (int i = 0; (i < que.size()) && (pf || qf); i++){if (que[i] == p)pf = false;if (que[i] == q)qf = false;qq.push_back(que[i]);qq1.push_back(que1[i]);if (que[i]->left != NULL){TN*n = new TN(que[i]->left->val);n->parent = que1[i];newque.push_back(que[i]->left);newque1.push_back(n);}if (que[i]->right != NULL){TN*n = new TN(que[i]->right->val);n->parent = que1[i];newque.push_back(que[i]->right);newque1.push_back(n);}}que = newque;que1 = newque1;}int pindex = find(qq.begin(), qq.end(), p) - qq.begin();int qindex = find(qq.begin(), qq.end(), q) - qq.begin();vector<TN*>pathp, pathq;TN*nn = qq1[pindex];while (nn!= NULL){pathp.push_back(nn);nn = nn->parent;}nn = qq1[qindex];while (nn != NULL){pathq.push_back(nn);nn = nn->parent;}int k = pathp.size()<pathq.size() ? pathp.size() : pathq.size();int i = 0;while (i<k&&pathp[pathp.size() - 1 - i] == pathq[pathq.size()-1-i])i++;if (i == k){TN*tt = pathp.size() < pathq.size() ? pathp[0] : pathq[0];int reindex = find(qq1.begin(), qq1.end(), tt) - qq1.begin();return qq[reindex];}else{TN*tt = pathp[pathp.size() - i];int reindex = find(qq1.begin(), qq1.end(), tt) - qq1.begin();return qq[reindex];}}};

accepted