[Google] LeetCode 2096 Step-By-Step Directions From a Binary Tree Node to Another

You are given the root of a binary tree with n nodes. Each node is uniquely assigned a value from 1 to n. You are also given an integer startValue representing the value of the start node s, and a different integer destValue representing the value of the destination node t.

Find the shortest path starting from node s and ending at node t. Generate step-by-step directions of such path as a string consisting of only the uppercase letters 'L', 'R', and 'U'. Each letter indicates a specific direction:

• 'L' means to go from a node to its left child node.
• 'R' means to go from a node to its right child node.
• 'U' means to go from a node to its parent node.

Return the step-by-step directions of the shortest path from node s to node t.

Solution

和 Oracle 有道题很像，那道题是求出最短路径的长度，这里让我们输出路径。

第一步依然是求出 \(LCA\). 很显然从 \(s\) 出发到 \(LCA\) 的路径必然是 \(U...\)，从 \(LCA\) 到 \(t\) 则是正常的 \(dfs\) 顺序即可。这里用 \(dfs\) 函数来判断当前的 \(v\) 是否存在 \(rt\) 为根的树里面。我们先将 \(L\) 加入到路径当中，如果不在该左子树里面，则 \(pop\_back\) 即可。

点击查看代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
private:
    string lca_to_p = "";
    string lca_to_q = "";
    
    TreeNode* LCA(TreeNode* rt, int p, int q){
        if(!rt) return NULL;
        if(rt->val==p || rt->val==q) return rt;
        TreeNode* lrt = LCA(rt->left, p, q);
        TreeNode* rrt = LCA(rt->right, p, q);
        if(lrt && rrt) return rt;
        else if(!lrt && !rrt) return NULL;
        return lrt?lrt:rrt;
    }
    
    bool dfs(TreeNode* rt, string& res, int v){
        if(!rt) return false;
        if(rt->val == v) return true;
        
        res.push_back('L');
        if(dfs(rt->left, res, v)){
            return true;
        }
        res.pop_back();
        res.push_back('R'); 
        if(dfs(rt->right, res, v)){
            return true;
        }
        res.pop_back();
        return false;
    }
    
    
public:
    string getDirections(TreeNode* root, int startValue, int destValue) {
        TreeNode* lca = LCA(root, startValue, destValue);
        
        dfs(lca, lca_to_p, startValue);
        dfs(lca, lca_to_q, destValue);
        
        for(auto &ele: lca_to_p){
            ele = 'U';
        }
        return lca_to_p+lca_to_q;
    }
};