Skip to content

Latest commit

 

History

History
156 lines (127 loc) · 3.33 KB

144.binary-tree-preorder-traversal.md

File metadata and controls

156 lines (127 loc) · 3.33 KB

题目地址

https://leetcode.com/problems/binary-tree-preorder-traversal/description/

题目描述

Given a binary tree, return the preorder traversal of its nodes' values.

Example:

Input: [1,null,2,3]
   1
    \
     2
    /
   3

Output: [1,2,3]
Follow up: Recursive solution is trivial, could you do it iteratively?

思路

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。

前序遍历是根左右的顺序,注意是开始,那么就很简单。直接先将根节点入栈,然后 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。

其他树的非递归遍历课没这么简单

关键点解析

  • 二叉树的基本操作(遍历)

    不同的遍历算法差异还是蛮大的

  • 如果非递归的话利用栈来简化操作

  • 如果数据规模不大的话,建议使用递归

  • 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

  1. 终止条件,自然是当前这个元素是 null(链表也是一样)

  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, 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), mid 是一个具体的节点,left 和 right递归求出即可

代码

  • 语言支持:JS,C++

JavaScript Code:

/*
 * @lc app=leetcode id=144 lang=javascript
 *
 * [144] Binary Tree Preorder Traversal
 *
 * https://leetcode.com/problems/binary-tree-preorder-traversal/description/
 *
 * algorithms
 * Medium (50.36%)
 * Total Accepted:    314K
 * Total Submissions: 621.2K
 * Testcase Example:  '[1,null,2,3]'
 *
 * Given a binary tree, return the preorder traversal of its nodes' values.
 *
 * Example:
 *
 *
 * Input: [1,null,2,3]
 * ⁠  1
 * ⁠   \
 * ⁠    2
 * ⁠   /
 * ⁠  3
 *
 * Output: [1,2,3]
 *
 *
 * Follow up: Recursive solution is trivial, could you do it iteratively?
 *
 */
/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[]}
 */
var preorderTraversal = function(root) {
  // 1. Recursive solution

  // if (!root) return [];

  // return [root.val].concat(preorderTraversal(root.left)).concat(preorderTraversal(root.right));

  // 2. iterative solutuon

  if (!root) return [];
  const ret = [];
  const stack = [root];
  let t = stack.pop();

  while (t) {
    ret.push(t.val);
    if (t.right) {
      stack.push(t.right);
    }
    if (t.left) {
      stack.push(t.left);
    }
    t = stack.pop();
  }

  return ret;
};

C++ Code:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> v;
        vector<TreeNode*> s;
        while (root != NULL || !s.empty()) {
            while (root != NULL) {
                v.push_back(root->val);
                s.push_back(root);
                root = root->left;
            }
            root = s.back()->right;
            s.pop_back();
        }
        return v;
    }
};