Tree Traversals

Tree Traversals

原题链接

常见的二叉树遍历的题目，根据后序遍历和中序遍历求层次遍历。

通过后序遍历和中序遍历建立起一棵二叉树，然后层序遍历一下，主要难点在于树的建立，通过中序遍历和后序遍历的特点递归求解，详细内容见代码

#include <iostream>
#include <queue>
using namespace std;

struct TreeNode{
    int val;
    TreeNode* left;
    TreeNode* right;
};
int post[];
int in[];
int n;
TreeNode* root;

void BuildTree(int l1, int r1, int l2, int r2, TreeNode*& root)
{
    root = new TreeNode();
    int i;
    for ( i = l2; i <= r2; i++)
    {
        if (in[i] == post[r1])
            break;
    }
    root->val = post[r1];
    if (i == l2)
        root->left = NULL;
    else
        BuildTree(l1, l1 + i - l2 - , l2, i - ,root->left); //边界条件
    if (i == r2)
        root->right = NULL;
    else
        BuildTree(l1+i-l2, r1 - , i + , r2, root->right); //边界条件

}

int flag = ;
void levelorder(TreeNode* T){
    queue<TreeNode*> q;
    q.push(T);
    while (!q.empty()){
        TreeNode* t = q.front();
        q.pop();
        if (t->left)
            q.push(t->left);
        if (t->right)
            q.push(t->right);
        if (flag == ){
            cout << t->val;
            flag = ;
        }
        else{
            cout << " " <<t->val;
        }
    }
    cout << endl;
}

void prevTree(TreeNode* root)
{
    if (root == NULL)
        return;
    cout << root->val << " ";
    prevTree(root->left);
    prevTree(root->right);
}

int main(void){
    cin >> n;
    for (size_t i = ; i < n; i++)
        cin >> post[i];
    for (size_t i = ; i < n; i++)
        cin >> in[i];

    BuildTree(, n - , , n - , root);

    levelorder(root);
}