[LeetCode] 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.

 

分析一：用两个辅助内存分别记录根节点到两个目标节点的路径。然后找出最后一个相同的节点。

　　      时间复杂度O(n), 空间复杂度一般O(lgn)，最差空间复杂度O(n)

/**
 * 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 || p == NULL || q == NULL)
            return NULL;
        
        vector<TreeNode*> path1;
        GetPath(root, p, path1);
        
        vector<TreeNode*> path2;
        GetPath(root, q, path2);
        
        return GetLastCommonNode(path1, path2);
    }
    
    TreeNode* GetLastCommonNode(const vector<TreeNode*>& path1, const vector<TreeNode*>& path2) {
        TreeNode* lastcommonnode = NULL;
        
        vector<TreeNode*>::const_iterator iter1 = path1.begin();
        vector<TreeNode*>::const_iterator iter2 = path2.begin();
        
        while (iter1 != path1.end() && iter2 != path2.end()) {
            if (*iter1 == *iter2) {
                lastcommonnode = *iter1;
            }
                
            iter1++;
            iter2++;
        }
        
        return lastcommonnode;
    }
    
    bool GetPath(TreeNode* root, TreeNode* p, vector<TreeNode*>& path) {
        path.push_back(root);
        
        if (root == p) return true;
        
        
        
        bool found = false;
        
        if (root->left != NULL) {
            found = GetPath(root->left, p, path);
        }
        
        if (!found && root->right != NULL)
            found = GetPath(root->right, p, path);
        
        if (!found)
            path.pop_back();
        
        return found;
    }
};

 

另一种方法：参考资料http://www.cnblogs.com/easonliu/p/4643873.html

/**
 * 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 (root == p || root == q) return root;
        TreeNode *L = lowestCommonAncestor(root->left, p, q);
        TreeNode *R = lowestCommonAncestor(root->right, p, q);
        if (L && R) return root;
        return L ? L : R;
    }
};