Lowest Common Ancestor of a Binary Tree Solutions in C++

// Source : https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/
// Author : Hao Chen
// Date   : 2015-07-17




/**
 * 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:
    bool findPath(TreeNode* root, TreeNode* p, vector<TreeNode*>& path) {
        if (root==NULL) return false;
        if (root == p) {
            path.push_back(p);
            return true;
        }
        
        path.push_back(root);
        if (findPath(root->left, p, path)) return true;
        if (findPath(root->right, p, path)) return true;
        path.pop_back();
        
        return false;
    }

    //Ordinary way, find the path and comapre the path.
    TreeNode* lowestCommonAncestor01(TreeNode* root, TreeNode* p, TreeNode* q) {
        
        vector<TreeNode*> path1, path2;
        
        if (!findPath(root, p, path1)) return NULL;
        if (!findPath(root, q, path2)) return NULL;
        
        int len = path1.size() < path2.size() ? path1.size() : path2.size();
        TreeNode* result = root;
        for(int i=0; i<len; i++) {
            if (path1[i] != path2[i]) {
                return result;
            }
            result = path1[i];
        }
        return result;
    }
    
    //Actually, we can do the recursive search in one time
    TreeNode* lowestCommonAncestor02(TreeNode* root, TreeNode* p, TreeNode* q) {
        
        //return if found or not found, return NULL if not found
        if (root==NULL || root == p || root == q) return root;
        
        //find the LCA in left tree
        TreeNode* left = lowestCommonAncestor02(root->left, p, q);
        //find the LCA in right tree
        TreeNode* right = lowestCommonAncestor02(root->right, p, q);
        
        //left==NULL means both `p` and `q` are not found in left tree.
        if (left==NULL) return right;
        //right==NULL means both `p` and `q` are not found in right tree.
        if (right==NULL) return left;
        // left!=NULL && right !=NULL, which means `p` & `q` are seperated in left and right tree.
        return root;
    }
    
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        srand(time(0));
        if (random()%2) {
            return lowestCommonAncestor02(root, p, q);
        }
        return lowestCommonAncestor01(root, p, q);
    }
};

/// Source : https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/description/
/// Author : liuyubobobo
/// Time   : 2020-04-22

#include <iostream>
#include <vector>

using namespace std;

/// Find two paths
/// Time Complexity: O(3n)
/// Space Complexity: O(2n)

///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) {

        vector<TreeNode*> path1;
        dfs(root, p, path1);

        vector<TreeNode*> path2;
        dfs(root, q, path2);

        TreeNode* res;
        for(int i = 0; i < path1.size() && i < path2.size(); i ++)
            if(path1[i] == path2[i]) res = path1[i];
            else break;
        return res;
    }

private:
    bool dfs(TreeNode* node, TreeNode* target, vector<TreeNode*>& path){

        if(!node) return false;

        path.push_back(node);
        if(node == target) return true;

        if(dfs(node->left, target, path)) return true;
        if(dfs(node->right, target, path)) return true;

        path.pop_back();
        return false;
    }
};


int main() {

    return 0;
}

/// Source : https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/description/
/// Author : liuyubobobo
/// Time   : 2017-01-30

#include <iostream>
#include <cassert>

using namespace std;

/// Recursion implementation
/// Time Complexity: O(n)
/// Space Complexity: 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:
    // 在root中寻找p和q
    // 如果p和q都在root所在的二叉树中, 则返回LCA
    // 如果p和q只有一个在root所在的二叉树中, 则返回p或者q
    // 如果p和q均不在root所在的二叉树中, 则返回NULL
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {

        if(root == NULL)
            return root;

        if(root == p || root == q)
            return root;

        TreeNode *left = lowestCommonAncestor(root->left, p, q);
        TreeNode *right = lowestCommonAncestor(root->right, p, q);

        if(left != NULL && right != NULL)
            return root;

        if(left != NULL)
            return left;

        if(right != NULL)
            return right;

        return NULL;
    }
};


int main() {

    return 0;
}

/// Source : https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/description/
/// Author : liuyubobobo
/// Time   : 2018-11-17

#include <iostream>
#include <cassert>

using namespace std;

/// Another Recursion implementation
/// Time Complexity: O(n)
/// Space Complexity: 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 {

private:
    TreeNode* ret = 0;

public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        dfs(root, p, q);
        assert(ret);
        return ret;
    }

private:
    // 在root中寻找p和q, 如果包含则返回 true
    // 否则返回false
    //
    // root是p或者q；root的左子树包含p或q；root的右子树包含p或q；三个条件有两个满足
    // 则 ret = root
    bool dfs(TreeNode* root, TreeNode* p, TreeNode* q) {

        if(!root)
            return false;

        bool mid = false;
        if(root == p || root == q)
            mid = true;

        bool left = dfs(root->left, p, q);
        bool right = dfs(root->right, p, q);

        if(mid + left + right >= 2)
            ret = root;

        return mid + left + right;
    }
};


int main() {

    return 0;
}

/// Source : https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/description/
/// Author : liuyubobobo
/// Time   : 2020-04-22

#include <iostream>
#include <vector>
#include <map>
#include <cmath>

using namespace std;

/// Binary Lifting
/// Time Complexity: init: O(nlogn)
///                  query: O(logn)
/// Space Complexity: O(nlogn)

///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:
    map<TreeNode*, int> tin, tout;
    map<TreeNode*, TreeNode*> parents;
    vector<TreeNode*> nodes;
    map<TreeNode*, vector<TreeNode*>> up;

public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {

        int time = 0;

        int depth = dfs(root, root, time);

        if(isAncestor(p, q)) return p;
        if(isAncestor(q, p)) return q;

        // up
        for(TreeNode* node: nodes)
            up[node].push_back(parents[node]);

        int d = ceil(log2(depth));
        for(int i = 1; i <= d; i ++)
            for(TreeNode* node: nodes)
                up[node].push_back(up[up[node][i - 1]][i - 1]);

        // lca
        for(int i = d; i >= 0; i --)
            if(!isAncestor(up[p][i], q))
                p = up[p][i];
        return up[p][0];
    }

private:
    // a is x's ancestor?
    bool isAncestor(TreeNode* a, TreeNode* x){
        return tin[a] <= tin[x] && tout[a] >= tout[x];
    }

    int dfs(TreeNode* node, TreeNode* p, int& time){

        if(!node) return 0;

        nodes.push_back(node);
        parents[node] = p;

        int depth = 0;

        tin[node] = time ++;
        depth = max(depth, 1 + dfs(node->left, node, time));
        depth = max(depth, 1 + dfs(node->right, node, time));
        tout[node] = time ++;

        return depth;
    }
};


int main() {

//    [-1,0,3,-2,4,null,null,8]
//    8
//    4
//    Expected: 0

    return 0;
}