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This approach uses a hash map to create nodes and establish their parent-child relationships. The root node is determined by finding nodes that haven't been a child in any relationship.
Time Complexity: O(N), where N is the number of descriptions.
Space Complexity: O(V), where V is the number of unique nodes.
1class TreeNode {
2 constructor(val, left = null, right = null) {
3 this.val = val;
4 this.left = left;
5 this.right = right;
6 }
7}
8
9function createBinaryTree(descriptions) {
10 const nodeMap = new Map();
11 const children = new Set();
12
13 for (const [parent, child, isLeft] of descriptions) {
14 if (!nodeMap.has(parent)) {
15 nodeMap.set(parent, new TreeNode(parent));
16 }
17 if (!nodeMap.has(child)) {
18 nodeMap.set(child, new TreeNode(child));
19 }
20
21 if (isLeft === 1) {
22 nodeMap.get(parent).left = nodeMap.get(child);
23 } else {
24 nodeMap.get(parent).right = nodeMap.get(child);
25 }
26 children.add(child);
27 }
28
29 for (const [key, node] of nodeMap) {
30 if (!children.has(key)) {
31 return node;
32 }
33 }
34 return null;
35}
36
The JavaScript solution uses a Map to keep track of each TreeNode instance along with their values. A Set identifies nodes that serve as children. The tree is constructed through relationship mapping, and the root node is determined by checking which node hasn't been listed as a child.
This approach uses a two-pass algorithm: the first pass creates all nodes independently, and the second pass links them based on the descriptive relationships. This avoids simultaneous creation and linking, providing a clearer separation of concerns between node creation and linkage.
Time Complexity: O(N), where N is the number of descriptions.
Space Complexity: O(V), where V is the number of unique nodes.
#include <unordered_set>
#include <vector>
using namespace std;
class TreeNode {
public:
int val;
TreeNode* left;
TreeNode* right;
TreeNode(int value) : val(value), left(nullptr), right(nullptr) {}
};
class Solution {
public:
TreeNode* createBinaryTree(vector<vector<int>>& descriptions) {
unordered_map<int, TreeNode*> nodes;
unordered_set<int> children;
// First pass to create all tree nodes
for (const auto& desc : descriptions) {
if (nodes.find(desc[0]) == nodes.end())
nodes[desc[0]] = new TreeNode(desc[0]);
if (nodes.find(desc[1]) == nodes.end())
nodes[desc[1]] = new TreeNode(desc[1]);
children.insert(desc[1]);
}
// Second pass to link tree nodes
for (const auto& desc : descriptions) {
int parent = desc[0], child = desc[1], isLeft = desc[2];
if (isLeft) {
nodes[parent]->left = nodes[child];
} else {
nodes[parent]->right = nodes[child];
}
}
for (const auto& [key, node] : nodes) {
if (children.find(key) == children.end()) {
return node;
}
}
return nullptr;
}
};
The C++ solution uses a two-pass algorithm similar to the C solution. The first pass initializes all nodes and records any child nodes. In the second pass, it establishes the prescribed left or right child relationships between the nodes.