[LeetCode]94.Binary Tree Inorder Traversal

Posted 2020-10-01 wzjhoutai

【题目】

Given a binary tree, return the inorder traversal of its nodes‘ values.

For example:
Given binary tree {1,#,2,3},

   1
         2
    /
   3

return [1,3,2].

Note: Recursive solution is trivial, could you do it iteratively?

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

【代码】

/*********************************
*   日期：2014-11-17
*   作者：SJF0115
*   题号: Binary Tree Inorder Traversal
*   来源：https://oj.leetcode.com/problems/binary-tree-inorder-traversal/
*   结果：AC
*   来源：LeetCode
*   总结：
**********************************/
#include <iostream>
#include <malloc.h>
#include <vector>
#include <stack>
using namespace std;

struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

class Solution {
public:
    vector<int> inorderTraversal(TreeNode *root) {
        vector<int> v;
        if (root == NULL){
            return v;
        }
        // 根节点入栈
        stack<TreeNode*> stack;
        TreeNode* node = root;
        // 遍历
        while(node != NULL || !stack.empty()){
            //遍历左子树
            if(node != NULL){
                stack.push(node);
                node = node->left;
            }
            else{
                //左子树为空，訪问右子树
                node = stack.top();
                stack.pop();
                v.push_back(node->val);
                node = node->right;
            }
        }
        return v;
    }
};

//按先序序列创建二叉树
int CreateBTree(TreeNode* &T){
    char data;
    //按先序次序输入二叉树中结点的值（一个字符），‘#’表示空树
    cin>>data;
    if(data == '#'){
        T = NULL;
    }
    else{
        T = (TreeNode*)malloc(sizeof(TreeNode));
        //生成根结点
        T->val = data-'0';
        //构造左子树
        CreateBTree(T->left);
        //构造右子树
        CreateBTree(T->right);
    }
    return 0;
}

int main() {
    Solution solution;
    TreeNode* root(0);
    CreateBTree(root);
    vector<int> v = solution.inorderTraversal(root);
    for(int i = 0;i < v.size();i++){
        cout<<v[i]<<endl;
    }
}