[LeetCode] Lowest Common Ancestor of a Binary Tree

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.

解题思路：

这道题的意思是求出二叉树任何两点的最低公共节点。与二分查找树的不同是，这个查找并没有规律，只能暴力法进行。

解法一：

递归法。倘若p, q都在root的左孩子树中，则令root=root->left。倘若p, q都在root的右孩子树种，则令root=root->right。否则，返回root。这种方法会有很多重复计算，产生超时错误。

/**
 * 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==NULL){
             return NULL;
         }
         if(isInTree(root->left, p) && isInTree(root->left, q)){
             return lowestCommonAncestor(root->left, p, q);
         }else if(isInTree(root->right, p) && isInTree(root->right, q)){
             return lowestCommonAncestor(root->right, p, q);
         }else{
             return root;
         }
    }
    
    bool isInTree(TreeNode* root, TreeNode* p){
        if(root==NULL){
            return false;
        }
        return root == p || isInTree(root->left, p) || isInTree(root->right, p);
    }
};

解法二：

可以先求出p，q到根节点的路径，然后找到路径中第一个相同的节点。

/**
 * 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==NULL){
            return NULL;
        }
        if(p==q){
            return p;
        }
        vector pPath;
        vector qPath;
        getPath(root, p, pPath);
        getPath(root, q, qPath);
        int i = pPath.size() - 1;
        int j = qPath.size() - 1;
        while(i>=0 && j>=0){
            if(pPath[i] == qPath[j]){
                i--;
                j--;
            }else{
                return pPath[i+1];
            }
        }
        
        return pPath[i+1];
    }
    
    bool getPath(TreeNode* root, TreeNode* p, vector& pPath){
        if(root==NULL){
            return false;
        }
        if(root==p){
            pPath.push_back(root);
            return true;
        }
        if(getPath(root->left, p, pPath) || getPath(root->right, p, pPath)){
            pPath.push_back(root);
            return true;
        }
        return false;
    }
};