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This approach leverages the recursive nature of trees to calculate the depth. For each node, if it has no children, it is a leaf and has depth 1. Otherwise, recursively calculate the depth of each child and take the maximum depth found among all children, adding 1 for the current node to account for the path to parent.
The time complexity is O(n), where n is the number of nodes in the tree. Each node is visited once. The space complexity is O(h) where h is the height of the tree, representing the function call stack during the recursion.
1class Node {
2 constructor(val, children = []) {
3 this.val = val;
4 this.children = children;
5 }
6}
7
8function maxDepth(root) {
9 if (!root) return 0;
10 if (root.children.length === 0) return 1;
11 let maxChildDepth = 0;
12 for (let child of root.children) {
13 maxChildDepth = Math.max(maxChildDepth, maxDepth(child));
14 }
15 return 1 + maxChildDepth;
16}In this JavaScript solution, a recursive approach similar to the Python solution is used. We iterate over each child and calculate their maximum depth recursively, updating the maximum depth found. We add 1 to the result to include the current node.
This approach utilizes BFS using a queue to iteratively compute tree depth. Nodes are enqueued level by level. At each level, we count its nodes, dequeuing them and enqueueing their children, indicating traversing to the next tree level. We increment a level counter as we progress deeper into the tree.
The time complexity is O(n) as we process each node once. The space complexity is O(n) for holding nodes of the widest level in the queue.
In this Java solution, BFS is applied iteratively. Nodes of each level are processed by polling the queue, and their children are enqueued. We increment the depth counter at each completed level to determine the tree's depth.