【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.

思路：

　　通过先序遍历，分别找到要查找结点（5 , 4）的路径（3->5 , 3->5->2->4），然后求路径序列中最后一个公共结点（5）即为所求。

C++:

 /**
  * 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 {
 private:
     vector<TreeNode* > plist;

 public:
     void rec(TreeNode *root, TreeNode *tar, vector<TreeNode* >& res)
     {
         if(root == tar)
         {
             vector<TreeNode* >::iterator it = plist.begin();
             for(; it != plist.end(); it++)
                 res.push_back(*it);

             return ;
         }

         if(root->left != )
         {
             plist.push_back(root->left);
             rec(root->left, tar, res);
             plist.pop_back();
         }

         if(root->right != )
         {
             plist.push_back(root->right);
             rec(root->right, tar, res);
             plist.pop_back();
         }
     }

     TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
         if(root == )
             return ;

         vector<TreeNode* > list1, list2;

         plist.push_back(root);
         rec(root, p, list1);

         plist.clear();
         plist.push_back(root);
         rec(root, q, list2);

         TreeNode  *ret = ;
         vector<TreeNode* >::iterator it1 = list1.begin();
         vector<TreeNode* >::iterator it2 = list2.begin();
         for(; it1 != list1.end() && it2 != list2.end(); it1++, it2++)
         {
             if(*it1 == *it2)
                 ret = *it1;
         }

         return ret;
     }
 };