236. Lowest Common Ancestor of a Binary Tree(最近公共祖先)

1 问题描述

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).”

Given the following binary tree: root = [3,5,1,6,2,0,8,null,null,7,4]

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.

Note:

• All of the nodes' values will be unique.
• p and q are different and both values will exist in the binary tree.

2 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) {
        vector<TreeNode*> path;
        vector<TreeNode*> p_node;
        vector<TreeNode*> q_node;
        int finish = 0;
        preorder(root, p, path, p_node, finish);
        path.clear();
        finish = 0;
        preorder(root, q, path, q_node, finish);
        int min_path_len = 0;
        if(p_node.size() < q_node.size())
            min_path_len = p_node.size();
        else
            min_path_len = q_node.size();

        TreeNode* result = 0;
        for(int i=0; i<min_path_len; i++) {
            if(p_node[i] == q_node[i]) {
                result = p_node[i];
            }
        }

        return result;
    }
private:
    void preorder(TreeNode* node, TreeNode* search, vector<TreeNode*> &path, vector<TreeNode*> &result, int finish) {
        if(!node || finish)
            return;

        path.push_back(node);

        if(node == search) {
            finish = true;
            result = path;
        }

        preorder(node->left, search, path, result, finish);
        preorder(node->right, search, path, result, finish);
        path.pop_back();
    }
};