655. 输出二叉树

https://leetcode.cn/problems/print-binary-tree/

解题思路

DFS

• 首先DFS获得深度，由此对于达不到深度的节点给构造子节点继续遍历
• 构造整个树为完全二叉树，所有节点都有左右子节点，达到最深节点的深度，如此方便构造数组
• 数组构造即左右子树的数组中间加一列空的“”，再加上当前节点的这一行

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 {
public:
    int getDeep(TreeNode* root, int n){
        // 获取深度
        if(!root) return n;
        return max(getDeep(root->left, n+1), getDeep(root->right, n+1));
    }

    vector<vector<string>> printTreeHelp(TreeNode* root, int deep, int level) {
        // deep 表示整个树深度， level表示当前深度为第几层
        vector<vector<string>> ans;
        if(level == deep) return ans; // 递归到最大深度
        vector<string> top;
        ans.push_back(top);
        string val;
        if(!root){ // 若当前节点为空，则值为空即可
            root = new TreeNode;
            val = "";
        }
        else val = to_string(root->val);

        vector<vector<string>> left = printTreeHelp(root->left, deep, level + 1);
        vector<vector<string>> right = printTreeHelp(root->right, deep, level + 1);
        int i = 0;

        // 拼凑左右子树数组
        for(; i < min(left.size(),right.size()); i++){
            vector<string> temp(left[i]); // 直接用左子树的该行数组初始化
            temp.push_back(""); // 拼凑时记得root这一列都有空字符串
            for(int j = 0; j < right[i].size(); j++){
                temp.push_back(right[i][j]);
            }
            ans.push_back(temp);
        }

        // 生成root节点的数组
        if(level == deep-1){ // 是最底层了，无子节点，直接返回
            ans[0].push_back(val);
            return ans;
        }
        for(int i = 0; i < ans[1].size(); i++){
            ans[0].push_back("");
        }
        ans[0][ans[1].size()/2] = val;
        return ans;
    }

    vector<vector<string>> printTree(TreeNode* root) {
        int deep = getDeep(root, 0);
        vector<vector<string>> ans;
        return printTreeHelp(root, deep, 0);
    }
};
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