157 lines
3.3 KiB
Markdown
157 lines
3.3 KiB
Markdown
## 题目地址
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https://leetcode.com/problems/binary-tree-preorder-traversal/description/
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## 题目描述
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```
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Given a binary tree, return the preorder traversal of its nodes' values.
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Example:
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Input: [1,null,2,3]
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1
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\
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2
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/
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3
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Output: [1,2,3]
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Follow up: Recursive solution is trivial, could you do it iteratively?
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```
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## 思路
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这道题目是前序遍历，这个和之前的`leetcode 94 号问题 - 中序遍历`完全不一回事。
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前序遍历是`根左右`的顺序，注意是`根`开始，那么就很简单。直接先将根节点入栈，然后
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看有没有右节点，有则入栈，再看有没有左节点，有则入栈。 然后出栈一个元素，重复即可。
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> 其他树的非递归遍历课没这么简单
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## 关键点解析
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- 二叉树的基本操作（遍历）
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> 不同的遍历算法差异还是蛮大的
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- 如果非递归的话利用栈来简化操作
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- 如果数据规模不大的话，建议使用递归
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- 递归的问题需要注意两点，一个是终止条件，一个如何缩小规模
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1. 终止条件，自然是当前这个元素是 null（链表也是一样）
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2. 由于二叉树本身就是一个递归结构， 每次处理一个子树其实就是缩小了规模，
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难点在于如何合并结果，这里的合并结果其实就是`mid.concat(left).concat(right)`,
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mid 是一个具体的节点，left 和 right`递归求出即可`
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## 代码
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- 语言支持：JS，C++
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JavaScript Code:
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```js
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/*
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* @lc app=leetcode id=144 lang=javascript
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*
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* [144] Binary Tree Preorder Traversal
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*
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* https://leetcode.com/problems/binary-tree-preorder-traversal/description/
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*
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* algorithms
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* Medium (50.36%)
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* Total Accepted: 314K
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* Total Submissions: 621.2K
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* Testcase Example: '[1,null,2,3]'
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*
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* Given a binary tree, return the preorder traversal of its nodes' values.
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*
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* Example:
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*
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*
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* Input: [1,null,2,3]
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* 1
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* \
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* 2
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* /
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* 3
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*
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* Output: [1,2,3]
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*
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*
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* Follow up: Recursive solution is trivial, could you do it iteratively?
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*
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*/
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/**
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* Definition for a binary tree node.
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* function TreeNode(val) {
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* this.val = val;
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* this.left = this.right = null;
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* }
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*/
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/**
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* @param {TreeNode} root
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* @return {number[]}
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*/
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var preorderTraversal = function(root) {
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// 1. Recursive solution
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// if (!root) return [];
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// return [root.val].concat(preorderTraversal(root.left)).concat(preorderTraversal(root.right));
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// 2. iterative solutuon
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if (!root) return [];
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const ret = [];
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const stack = [root];
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let t = stack.pop();
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while (t) {
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ret.push(t.val);
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if (t.right) {
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stack.push(t.right);
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}
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if (t.left) {
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stack.push(t.left);
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}
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t = stack.pop();
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}
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return ret;
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};
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```
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C++ Code:
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```C++
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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* };
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*/
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class Solution {
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public:
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vector<int> preorderTraversal(TreeNode* root) {
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vector<int> v;
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vector<TreeNode*> s;
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while (root != NULL || !s.empty()) {
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while (root != NULL) {
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v.push_back(root->val);
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s.push_back(root);
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root = root->left;
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}
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root = s.back()->right;
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s.pop_back();
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}
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return v;
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}
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};
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```
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