# 面试题68 - II

给定一个二叉树, 找到该树中两个指定节点的最近公共祖先。

百度百科中最近公共祖先的定义为：“对于有根树 T 的两个结点 p、q，最近公共祖先表示为一个结点 x，满足 x 是 p、q 的祖先且 x 的深度尽可能大（一个节点也可以是它自己的祖先）。”

例如，给定如下二叉树: root = \[3,5,1,6,2,0,8,null,null,7,4]

```
示例 1:

输入: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
输出: 3
解释: 节点 5 和节点 1 的最近公共祖先是节点 3。

示例 2:

输入: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
输出: 5
解释: 节点 5 和节点 4 的最近公共祖先是节点 5。因为根据定义最近公共祖先节点可以为节点本身。
```

## 说明:

* 所有节点的值都是唯一的。
* p、q 为不同节点且均存在于给定的二叉树中。

## 注意：本题与主站 236 题相同：[https://leetcode-cn.com/problems/lowest-common-ancestor-of-a-binary-tree/](https://leetcode-cn.com/problems/lowest-common-ancestor-of-a-binary-tree/)

## Solutions

1. **recursion**

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

1. **hashmap**
2. Record each nodes' parent node.
3. Then find the first common ancestor of two given nodes by traversing the ancester tree of each node.

```cpp
```

1. **inorder traversal with stack**

```cpp
class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        stack<TreeNode *> s;
        TreeNode * lca = nullptr;
        int find = 0, lcalevel = INT_MAX;

        while (root || !s.empty()) {
            if (root) {
                s.push(root);
                root = root->left;
            }
            else {
                root = s.top(); s.pop();
                if (root == p || root == q)
                    find++;
                if (find && s.size() < lcalevel) {
                    lca = root;
                    lcalevel = s.size();
                }
                if (find == 2)
                    return lca;

                root = root->right;
            }
        }

        return nullptr;
    }
};
```
