lintcode lowest-common-ancestor 最近公共祖先

问题描述

lintcode

笔记

先分别找到两个节点的路径，然后对比两条路径。

代码

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */
class Solution {
private:
    vector pathA;
    vector pathB;
    vector<vector > pathArr;
public:
    /**
     * @param root: The root of the binary search tree.
     * @param A and B: two nodes in a Binary.
     * @return: Return the least common ancestor(LCA) of the two nodes.
     */
    TreeNode *lowestCommonAncestor(TreeNode *root, TreeNode *A, TreeNode *B) {
        // write your code here
        pathA.clear();
        pathB.clear();
        pathArr.clear();
        int findA = 0;
        int findB = 0;
        getPath(root, A, pathA, pathArr);
        getPath(root, B, pathB, pathArr);
        TreeNode *res = NULL;
        if (pathArr[0].empty() || pathArr[1].empty())
        {
            return res;
        }
        else
        {
            int i = 0;
            int lenA = pathArr[0].size();
            int lenB = pathArr[1].size();
            // cout << "lena lenb "<< lenA << " " << lenB << " ";
            while (i < lenA && i < lenB && pathArr[0][i] == pathArr[1][i])
            {
                i++;
            }
            return pathArr[0][i-1];

        }
    }

    void getPath(TreeNode *root, TreeNode *A, vector &path, vector<vector > &res)
    {
        if (root == NULL)
        {
            // printf("\nfindA=%d\n", findA);
            return;
        }
        path.push_back(root);
        if (root == A)
        {
            res.push_back(path);
        }
        if (root->left)
            getPath(root->left, A, path, pathArr);
        if (root->right)
            getPath(root->right, A, path, pathArr);
        path.pop_back();
    }
};