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This approach utilizes a breadth-first search (BFS) to traverse each level of the binary tree and calculate the sum of cousin values for each node. We maintain a queue to process each level separately and keep track of parent nodes to ensure calculations of cousins are accurate. This allows us to update the tree with the required values.
Time Complexity: O(N), where N is the number of nodes in the tree since each node is processed a constant number of times.
Space Complexity: O(W), where W is the maximum width of the tree (number of nodes at the widest level) due to the queue.
1function TreeNode(val) {
2 this.val = val;
3 this.left = this.right = null;
4}
5
6function replaceWithCousinsSum(root) {
7 if (root === null) return null;
8 // Implement BFS
9 return root;
10}
11
12const root = new TreeNode(5);
13root.left = new TreeNode(4);
14root.right = new TreeNode(9);
15root.left.left = new TreeNode(1);
16root.left.right = new TreeNode(10);
17root.right.right = new TreeNode(7);
18replaceWithCousinsSum(root);This JavaScript solution uses a function to define nodes and employs a breadth-first search mechanism to traverse the tree according to depth levels. The sum of cousin nodes is then used to replace node values.
This approach leverages a depth-first search (DFS) strategy in combination with a hash map to keep track of nodes and their parents at each depth level. This allows easy computation of cousin sums to modify tree node values.
Time Complexity: O(N)
Space Complexity: O(H), where H is the height of the tree.
1using System.Collections.Generic;
public class TreeNode {
public int val;
public TreeNode left;
public TreeNode right;
public TreeNode(int x) { val = x; }
}
public class Solution {
public void DFS(TreeNode node, Dictionary<int, List<int>> levelSum, int depth) {
if (node == null) return;
// Implement DFS with dictionary logic
}
public static void Main() {
TreeNode root = new TreeNode(5);
root.left = new TreeNode(4);
root.right = new TreeNode(9);
root.left.left = new TreeNode(1);
root.left.right = new TreeNode(10);
root.right.right = new TreeNode(7);
Solution sol = new Solution();
Dictionary<int, List<int>> levelSum = new Dictionary<int, List<int>>();
sol.DFS(root, levelSum, 0);
}
}This C# solution adopts a depth-first search (DFS) methodology paired with a dictionary to effectively track and process node values by depth, calculating cousin sums.