在读leetcode第337题时，发现题目中的二叉树排列的很规整、漂亮，如下：
示例一： 输入: [3,2,3,null,3,null,1] 3 / \ 2 3 \ \ 3 1 输出: 7 示例二： 输入: [3,4,5,1,3,null,1] 3 / \ 4 5 / \ \ 1 3 1 输出: 9于是参考了博客，并把格式加以修改，最后在Vs 2019 中打印得到了该树形，完整代码详见链接或者文末。需要注意的是第二行的叶子结点不能同时拥有右孩子和左孩子，打印示意图如下:
完整代码如下：
#include <iostream> #include <iomanip> #include <cmath> #include <string> #include <vector> using namespace std; struct TreeNode { //树结点定义 int val; TreeNode* left; TreeNode* right; TreeNode(int x) : val(x), left(NULL), right(NULL) {} }; class Solution { public: void read(int input[], int& num); //读入输入函数 TreeNode* buildTree(int input[], int& num); //生成二叉树函数 }; void Solution::read(int input[], int& num) { string buffer; cin >> buffer; //输入 string::iterator ster = buffer.begin(); for (; ster < buffer.end(); ++ster) { if (*ster == '[' || *ster == ',') continue; else if (*ster == ']') break; else if (*ster != 'n') input[num] = atoi(&(*ster)); else { input[num] = -1; //如果为null,val赋值为-1 ster += 4; } ++num; //指向下一个 } --num; //指向末尾 } TreeNode* Solution::buildTree(int input[], int& num) { TreeNode* treenode[32]; //存入树节点的指针 for (int i = 1; i <= num; ++i) { TreeNode* temp = new TreeNode(input[i]); //生成树节点 treenode[i] = temp; //记录指向结点的指针，方便连接树结构 if (i != 1) { //不是根节点 if (i % 2 == 0) { treenode[i / 2]->left = temp; //该结点为其根节点的左孩子 } else { treenode[i / 2]->right = temp; //该结点为其根节点的右孩子 } } } return treenode[1]; //返回根节点 } //树形打印函数 int getHeight(const TreeNode* node) //二叉树的高度 { int treeHeight = 0; if (node != NULL) { int leftHeight = getHeight(node->left); int rightHeight = getHeight(node->right); treeHeight = leftHeight >= rightHeight ? leftHeight + 1 : rightHeight + 1; } return treeHeight; } void getLine(const TreeNode* root, int depth, vector<int>& vals) { int placeholder = -1; if (depth <= 0 && root != NULL) { vals.push_back(root->val); return; } if (root->left != NULL) getLine(root->left, depth - 1, vals); else if (depth - 1 <= 0) vals.push_back(placeholder); if (root->right != NULL) getLine(root->right, depth - 1, vals); else if (depth - 1 <= 0) vals.push_back(placeholder); } int flag = 0; void printRow(const TreeNode* p, int depth) { vector<int> vec; int placeholder = -1; getLine(p, depth, vec); cout << setw(flag--); // scale setw with depth bool toggle = true; // start with left if (vec.size() > 1) { for (int v = 0; v < vec.size(); ++v) { if (vec[v] != placeholder) { if (toggle) cout << "/"; else cout << " \\" << setw(2 * (depth - 1)); } else { if (toggle) cout << " "; else cout << " "; } toggle = !toggle; } cout << endl; cout << setw(flag--); //输出宽度递减，为输出叶子做准备 } toggle = true; // start with left for (int v = 0; v < vec.size(); v++) { if (vec[v] != -1) { if (toggle) cout << vec[v] << " "; //第一个空格占位 else cout << vec[v] << setw(pow(2, flag + 1) - 1); } else { if (v == 0) cout << " "; //第一个空格占位 if (toggle) cout << " " << setw(3); } toggle = !toggle; } cout << endl; } void postorder(TreeNode* p) { int height = getHeight(p); flag = height + 2; //取齐第一次减减 for (int i = 0; i < getHeight(p); i++) { printRow(p, i); height--; } } int main() { int num = 1, input[32]; //input 存入结点值 从1开始存入, num 记录结点数 Solution solu; solu.read(input, num); //读入数据，并加以处理 TreeNode* root = solu.buildTree(input, num); //生成二叉树 postorder(root); }参考： https://blog.csdn.net/Diano_lx/article/details/85764548
