题目：

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 p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

 

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.

Example 2:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

Example 3:

Input: root = [1,2], p = 1, q = 2
Output: 1

 

Constraints:

• The number of nodes in the tree is in the range [2, 105].
• -109 <= Node.val <= 109
• All Node.val are unique.
• p != q
• p and q will exist in the tree.

 

代码：

又是一题二叉树~

/**
 * 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:
    void findNode(TreeNode* root, TreeNode* p, vector<TreeNode*>& path) {
        if(root == NULL)
            return;
        if(root->val == p->val)
        {
            path.push_back(root);
            return;
        }
        if(root->left != NULL)
        {
            path.push_back(root);
            findNode(root->left, p, path);
            if(path[path.size()-1]->val != p->val)
                path.pop_back();
        }
        if(root->right != NULL)
        {
            path.push_back(root);
            findNode(root->right, p, path);
            if(path[path.size()-1]->val != p->val)
                path.pop_back();
        }
        return;
    }
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        vector<TreeNode*> p_path, q_path;
        findNode(root, p, p_path);
        findNode(root, q, q_path);
        TreeNode* ans = root;
        for(int i = 0; i<p_path.size() && i<q_path.size(); i++)
        {
            if(p_path[i] != q_path[i])
            {
                break;
            }
            ans = p_path[i];
        }
        return ans;
    }
};

 

新的一个月~加油吧！