Here is the **full question rewritten cleanly in t...

Below is an O(n) C++ solution (works for n ≤ 3e5) using a standard tree DP for perfect matching on every rooted subtree (root assumed to be node 1, matching the usual “subtree rooted at i” definition used in these problems and your example).

Key idea (tree DP)

Root the tree at 1. For each node u, compute:

• dp0[u] = subtree of u can be perfectly paired except u is left unpaired (so u would need to pair with its parent outside the subtree).
• dp1[u] = subtree of u can be perfectly paired and u is paired with exactly one of its children (fully paired inside the subtree).

For the answer for node u’s subtree (which does not include its parent), we need u paired inside the subtree, so output '1' iff dp1[u] is true.

Transitions:

• dp0[u] is true iff all children c satisfy dp1[c] (each child must be fully matched internally).
• dp1[u] is true iff there exists a child t such that:
  • dp0[t] is true (child t can be internally matched except itself, because it will match with u)
  • all other children c ≠ t satisfy dp1[c]

We compute the “all other children” condition efficiently with prefix/suffix AND in O(deg(u)).

✅ Correct C++ Code

cpp
#include <bits/stdc++.h>
using namespace std;

string isColoringPossible(
    int network_nodes,
    vector<int> network_from,
    vector<int> network_to
) {
    int n = network_nodes;
    int m = (int)network_from.size();

    // Build adjacency with degree-based reserve (faster + fewer reallocs)
    vector<int> deg(n + 1, 0);
    for (int i = 0; i < m; i++) {
        deg[network_from[i]]++;
        deg[network_to[i]]++;
    }

    vector<vector<int>> adj(n + 1);
    for (int i = 1; i <= n; i++) adj[i].reserve(deg[i]);

    for (int i = 0; i < m; i++) {
        int a = network_from[i], b = network_to[i];
        adj[a].push_back(b);
        adj[b].push_back(a);
    }

    // Root the tree at 1
    vector<int> parent(n + 1, 0);
    vector<vector<int>> children(n + 1);
    for (int i = 1; i <= n; i++) children[i].reserve(deg[i]);

    vector<int> order;
    order.reserve(n);

    vector<int> st;
    st.reserve(n);
    st.push_back(1);
    parent[1] = -1;

    while (!st.empty()) {
        int u = st.back();
        st.pop_back();
        order.push_back(u);

        for (int v : adj[u]) {
            if (v == parent[u]) continue;
            parent[v] = u;
            children[u].push_back(v);
            st.push_back(v);
        }
    }

    // dp0[u]: subtree(u) matched except u is free
    // dp1[u]: subtree(u) fully matched and u matched with one child
    vector<char> dp0(n + 1, 0), dp1(n + 1, 0);

    // Postorder compute
    for (int idx = n - 1; idx >= 0; --idx) {
        int u = order[idx];
        auto &ch = children[u];
        int k = (int)ch.size();

        if (k == 0) {
            dp0[u] = 1;   // leaf can be "all matched except itself"
            dp1[u] = 0;   // can't match inside (no child)
            continue;
        }

        // prefix/suffix AND of dp1 on children
        vector<char> pref(k + 1, 1), suf(k + 1, 1);
        for (int i = 0; i < k; i++) pref[i + 1] = (char)(pref[i] && dp1[ch[i]]);
        for (int i = k - 1; i >= 0; i--) suf[i] = (char)(suf[i + 1] && dp1[ch[i]]);

        // u is free => all children must be fully matched internally
        dp0[u] = pref[k];

        // u matched to exactly one child t:
        // child t must be dp0[t] (t free internally), others must be dp1
        char ok = 0;
        for (int i = 0; i < k; i++) {
            int t = ch[i];
            if (dp0[t] && pref[i] && suf[i + 1]) {
                ok = 1;
                break;
            }
        }
        dp1[u] = ok;
    }

    // Answer for each node i: subtree rooted at i must match i internally => dp1[i]
    string ans;
    ans.reserve(n);
    for (int i = 1; i <= n; i++) {
        ans.push_back(dp1[i] ? '1' : '0');
    }
    return ans;
}

Complexity

• Time: O(n)
• Memory: O(n)

If you want, I can also provide a version that avoids storing children (slightly lower memory), but the above is already safe and fast for 3e5.