Lowest Common Ancestor of a Binary Tree

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题目

https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/description/

-w953

想法

自己想的是，dfs找到两者的路径，然后比较两者相同的前缀。

不过，看了discussion中的一个解答之后，发现自己还真是naive了，看准情况，暴力的才是最好的！

答案

我的答案：

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/**
 * 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<int, vector<TreeNode*>> path;
    vector<TreeNode*> curr;
    void dfs(TreeNode* node, int k) {
        if (!node)
            return;
        curr.push_back(node);
        if (node->val == k) {
            path[k] = curr;
        } else {
            dfs(node->left, k);
            dfs(node->right, k);
        }
        curr.pop_back();
    }
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        dfs(root, p->val);
        dfs(root, q->val);
        TreeNode* res = root;
        int i = 0;
        while (true) {
            if (i >= path[p->val].size() || i >= path[q->val].size() || path[p->val][i] != path[q->val][i]) {
                break;
            } else {
                res = path[p->val][i];
                i++;
            }
        }
        return res;
    }
};

discussion中某位的答案：
https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/discuss/65225/4-lines-C++JavaPythonRuby

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TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
    if (!root || root == p || root == q) return root;
    TreeNode* left = lowestCommonAncestor(root->left, p, q);
    TreeNode* right = lowestCommonAncestor(root->right, p, q);
    return !left ? right : !right ? left : root;
}