LeetCode: 236. Lowest Common Ancestor of a Binary Tree

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题目描述

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]

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Example 1:

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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:

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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.

这题要在二叉树中寻找两个点的最低的ancestor。

思路就是用两次DFS，找到两个点p和点q，并且记录路径，最后找路径中最先出现的相同点就可以了。在找路径出现的相同点点时候，使用set来找。

代码实现

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class Solution {
private:
    int findNode(TreeNode* root, TreeNode* p, vector<TreeNode*>& path) {
        path.push_back(root);

        if (root == p) {
            return 1;
        }

        if (root->left != NULL){
            if (findNode(root->left, p, path))
                return 1;
            path.pop_back();
        }

        if (root->right != NULL){
            if (findNode(root->right, p, path))
                return 1;
            path.pop_back();
        }

        return 0;
    }

public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        set<TreeNode*> setNode;
        vector<TreeNode*> path1, path2;

        findNode(root, p, path1);
        findNode(root, q, path2);

        for (int i = 0; i < path1.size(); i++) {
            setNode.insert(path1[i]);
        }

        for (int i = path2.size()-1; i >= 0; i--) {
            if (setNode.find(path2[i])!=setNode.end())
                return path2[i];
        }
        return NULL;
    }
};