1099. Build A Binary Search Tree (30)

    从根结点开始，每次统计根节点左边孩子节点的数目n，以此确定当前根节点的key值，应该是所有的key排序之后的数组的第n+1个值，递归构建即可。

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <queue>

using namespace std;

struct Node{
int key, left, right;

Node(){}
Node(int k, int l, int r) : key(k), left(l), right(r){}
};

vector<Node> nodes;
vector<int> keys;

int count(int root){
if(root == -1) return 0;

return count(nodes[root].left) + count(nodes[root].right) + 1;
}

void fillKey(int root, int start){
if(root == -1) return;

int left = count(nodes[root].left);
nodes[root].key = keys[start+left];

fillKey(nodes[root].left, start);
fillKey(nodes[root].right, start+left+1);
}

int main(){
int n;
scanf("%d", &n);
nodes.resize(n);
keys.resize(n);

for(int i = 0; i < n; ++i){
int l, r;
scanf("%d%d", &l, &r);
nodes[i] = Node(0, l, r);
}

for(int i = 0; i < n; ++i){
scanf("%d", &keys[i]);
}

sort(begin(keys), end(keys));
fillKey(0, 0);

queue<Node> Q;
Q.push(nodes[0]);
vector<int> result;

while(!Q.empty()){
Node node = Q.front(); Q.pop();
result.push_back(node.key);

if(node.left != -1) Q.push(nodes[node.left]);
if(node.right != -1) Q.push(nodes[node.right]);
}

for(size_t i = 0; i < result.size(); ++i){
if(i) printf(" ");
printf("%d", result[i]);
}

return 0;
}