1099. Build A Binary Search Tree

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给出二叉树结构和节点的值列表, 将值填入二叉树中, 输出层次序遍历的结果

考点: 二叉树的直接后继

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#include <iostream>
#include <iomanip>
#include <algorithm>
#include <vector>
#include <string>
#include <queue>
#include <map>
#include <set>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <climits>
using namespace std;

#ifdef _DEBUG
template <class T> void ERR(T x, string name)
{
    cout << "[DEBUG]{" << name << "} = " << x << endl;
}
#    define debug(x) ERR(x, #    x)
#else
#    define debug(...) 0
#endif

struct node {
    int val;
    int left, right, parent;
} N[100];

int main()
{
#ifdef _DEBUG
    freopen(__FILE__ ".txt", "r", stdin);
    // freopen(__FILE__ ".out", "w", stdout);
#endif
    int first = 1;

    int n;
    cin >> n;
    N[0].parent = -1;
    for (int i = 0; i < n; i++) {
        int l, r;
        cin >> l >> r;
        N[i].left = l;
        N[l].parent = i;
        N[i].right = r;
        N[r].parent = i;
    }

    vector<int> vals;
    for (int i = 0; i < n; i++) {
        int val;
        cin >> val;
        vals.push_back(val);
    }
    sort(vals.begin(), vals.end());

    int p = 0;
    while (N[p].left != -1) {
        if (N[p].left != -1)
            p = N[p].left;
    }

    while (p != -1) {
        debug(p);
        N[p].val = vals.front();
        vals.erase(vals.begin());

        if (N[p].right != -1) {
            // right down
            p = N[p].right;
            while (N[p].left != -1) {
                p = N[p].left;
            }
        } else if (N[p].parent != -1 && N[N[p].parent].left == p) {
            // right up
            p = N[p].parent;
        } else {
            while (N[p].parent != -1 && N[N[p].parent].right == p) {
                p = N[p].parent;
            }
            p = N[p].parent;
        }
    }

    queue<int> Q;
    Q.push(0);
    while (Q.size()) {
        int top = Q.front();
        Q.pop();
        if (!first)
            cout << " ";
        else
            first = 0;
        cout << N[top].val;
        if (N[top].left != -1)
            Q.push(N[top].left);
        if (N[top].right != -1)
            Q.push(N[top].right);
    }

    return 0;
}