easy Maximum Depth or Height of Binary Tree
Problem Statement
Given a root node of the binary tree, return the depth (or height) of a binary tree.
The Depth of the binary tree refers to the number of nodes along the longest path from the root node to the farthest leaf node. If the tree is empty, the depth is 0.
Examples
Example 1
- Input: root = [1, 2, 3, 4, 5]
- Expected Output: 3
- Explanation: The longest path is
1->2->4or1->2->5with 3 nodes.
Example 2
- Input: root = [1, null, 2, null, 3]
- Expected Output: 3
- Justification: There's only one path
1->2->3with 3 nodes.
Example 3
- Input: root = [1, 2, 3, 4, 7, null, null, null, null, null, 9]
- Expected Output: 4
- Justification: The longest path is
1->2->7->9with 4 nodes.
Constraints:
- The number of nodes in the tree is in the range [0, 104].
-100 <= Node.val <= 100
Try it yourself
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✅ Solution Maximum Depth or Height of Binary Tree
Problem Statement
Given a root node of the binary tree, return the depth (or height) of a binary tree.
The Depth of the binary tree refers to the number of nodes along the longest path from the root node to the farthest leaf node. If the tree is empty, the depth is 0.
Examples
Example 1
- Input: root = [1, 2, 3, 4, 5]
- Expected Output: 3
- Explanation: The longest path is
1->2->4or1->2->5with 3 nodes.
Example 2
- Input: root = [1, null, 2, null, 3]
- Expected Output: 3
- Justification: There's only one path
1->2->3with 3 nodes.
Example 3
- Input: root = [1, 2, 3, 4, 7, null, null, null, null, null, 9]
- Expected Output: 4
- Justification: The longest path is
1->2->7->9with 4 nodes.
Constraints:
- The number of nodes in the tree is in the range [0, 104].
-100 <= Node.val <= 100
Solution
What's the 'Depth'?
Picture a ladder. Each step is a level in our tree. The depth? It's just how many steps there are! If we have a 3-step ladder, our tree has a depth of 3.
To determine the deepest level of a binary tree, we'll use a depth-first search (DFS) approach. Begin at the root node and traverse down each branch, keeping track of the current depth. The depth increases by one each time we move to a child node. If a node is a leaf (no children), compare its depth with the current maximum depth and update if necessary. Recursively apply this process to each node's left and right children. This will cover all paths in the tree, allowing us to find and return the maximum depth encountered.
Step-by-step Algorithm
-
Recursive Approach: The depth of a binary tree can be computed recursively. The maximum depth of a tree is the maximum of the depths of its left and right subtrees plus one (the root node).
-
Base Condition: If the node is null, return 0. This ensures that we've reached the end of a branch or the tree is empty.
-
Recursive Calculation: For each node, compute the depth of its left subtree and the depth of its right subtree. The depth of the current node is 1 + the maximum of these two depths.
-
Result: For the root node, this recursive approach will provide the maximum depth of the entire tree.
Algorithm Walkthrough
-
Starting at the root node
1:- The method
maxDepth(root)is called with the root node1. - The root node is not
null, so we proceed to calculate the depth of its left and right subtrees.
- The method
-
Calculating the depth of the left subtree of node
1:- The method
maxDepth(root.left)is called with the left child of node1, which is node2.
- The method
-
Calculating the depth of the left subtree of node
2:- The method
maxDepth(root.left.left)is called with the left child of node2, which is node4. - Node
4is a leaf node (no left or right children). - The method
maxDepth(root.left.left.left)is called withnull(left child of4), returning0. - The method
maxDepth(root.left.left.right)is called withnull(right child of4), returning0. - The maximum depth from node
4is1 + max(0, 0) = 1.
- The method
-
Calculating the depth of the right subtree of node
2:- The method
maxDepth(root.left.right)is called with the right child of node2, which is node5. - Node
5is a leaf node (no left or right children). - The maximum depth from node
5is1 + max(0, 0) = 1.
- The method
-
Determining the depth of node
2:- Node
2has a left depth of1(from node4) and a right depth of1(from node5). - The maximum depth from node
2is1 + max(1, 1) = 2.
- Node
-
Calculating the depth of the right subtree of node
1:- The method
maxDepth(root.right)is called with the right child of node1, which is node3. - Node
3is a leaf node (no left or right children). - The maximum depth from node
3is1 + max(0, 0) = 1.
- The method
-
Determining the depth of the root node
1:- Node
1has a left depth of2(from node2) and a right depth of1(from node3). - The maximum depth of the tree rooted at node
1is1 + max(2, 1) = 3.
- Node
Final Output: The maximum depth of the tree [1, 2, 3, 4, 5] is 3.
Code
Here is the code for this algorithm:
// Definition for a binary tree node.
// class TreeNode {
// int val;
// TreeNode left;
// TreeNode right;
// TreeNode(int x) { val = x; }
// }
public class Solution {
// Method to compute the maximum depth of a binary tree
public int maxDepth(TreeNode root) {
// Base case: if node is null, return 0
if (root == null) return 0;
// Recursively calculate left subtree depth
int leftDepth = maxDepth(root.left);
// Recursively calculate right subtree depth
int rightDepth = maxDepth(root.right);
// Return the maximum of left and right subtree depth plus 1 for the current node
return Math.max(leftDepth, rightDepth) + 1;
}
public static void main(String[] args) {
Solution solver = new Solution();
// Example 1
TreeNode root1 = new TreeNode(1);
root1.left = new TreeNode(2);
root1.right = new TreeNode(3);
root1.left.left = new TreeNode(4);
root1.left.right = new TreeNode(5);
System.out.println(solver.maxDepth(root1)); // Expected output: 3
// Example 2
TreeNode root2 = new TreeNode(1);
root2.right = new TreeNode(2);
root2.right.right = new TreeNode(3);
System.out.println(solver.maxDepth(root2)); // Expected output: 3
// Example 3
TreeNode root3 = new TreeNode(1);
root3.left = new TreeNode(2);
root3.right = new TreeNode(3);
root3.left.left = new TreeNode(4);
root3.left.right = new TreeNode(7);
root3.left.right.right = new TreeNode(9);
System.out.println(solver.maxDepth(root3)); // Expected output: 4
}
}
Complexity Analysis
Time Complexity
-
Recursive Calls:
- The algorithm uses recursion to traverse each node in the binary tree exactly once. For each node, it makes a recursive call for the left child and the right child.
- If the binary tree has
nnodes, there will benrecursive calls in total. - Therefore, the time complexity for traversing the entire tree is
.
-
Calculating Maximum Depth:
- For each node, the algorithm calculates the maximum of the left and right subtree depths, which is a constant time operation
.
- For each node, the algorithm calculates the maximum of the left and right subtree depths, which is a constant time operation
Combining these steps, the overall time complexity of the algorithm is
Space Complexity
-
Recursive Stack Space:
- The depth of the recursion stack is determined by the height of the tree. In the worst case, if the tree is skewed (i.e., all nodes are in a single line), the height of the tree will be
n, resulting in a space complexity offor the recursion stack. - In the best case (a balanced tree), the height of the tree will be
, resulting in a space complexity of .
- The depth of the recursion stack is determined by the height of the tree. In the worst case, if the tree is skewed (i.e., all nodes are in a single line), the height of the tree will be
-
Additional Space:
- The algorithm does not use any additional space other than the recursion stack.
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