面试题68 - II： 二叉树的最近公共祖先（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 {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(root == nullptr || root == p || root == q) return root; //返回最近公共祖先节点
        TreeNode* left = lowestCommonAncestor(root->left, p, q); //递归左子树
        TreeNode* right = lowestCommonAncestor(root->right, p, q); //递归右子树
        if(left == nullptr && right == nullptr) return nullptr; //case1
        if(left == nullptr) return right; //case3
        if(right == nullptr) return left; //case4
        return root; //case2
    }
};

/**
 * 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 path_p, path_q;
            node_path(root, p, path_p); //获取根节点root到节点p的路径
            node_path(root, q, path_q); //获取根节点root到节点q的路径
            TreeNode* node = nullptr; // 最近公共祖先节点
            int node_len = min(path_p.size(), path_q.size());
            for(int i = 0; i )
            {
                if(path_p[i] != path_q[i]) return node;
                node = path_p[i];
            }
            return node;
    }
    void node_path(TreeNode* root, TreeNode* last, vector& path)
    {
        stack sk;
        TreeNode* cur = root, *prev = nullptr; //prev表示当前节点已被访问
        while(cur != nullptr || !sk.empty())
        {
            while(cur)
            {
                sk.push(cur);
                path.push_back(cur);
                if(cur == last) return;
                cur = cur->left; //遍历左子树
            }
            cur = sk.top();
            if(cur->right == nullptr || cur->right == prev)
            {
                path.pop_back(); //清空路径
                sk.pop();
                prev = cur;
                cur = nullptr;
            }
            else
            {
                cur = cur->right; //遍历右子树
            }
        }
    }
};