104. Maximum Depth of Binary Tree
Given the root of a binary tree, return its maximum depth.
A binary tree's maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.
Example 1:
Example 2:
Recursion
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public int maxDepth(TreeNode root) {
if (root == null){
return 0;
}
int left = maxDepth(root.left);
int right = maxDepth(root.right);
return Math.max(left, right) + 1;
}
}
// TC: O(n)
// SC: O(n)
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public int maxDepth(TreeNode root) {
if (root == null){
return 0;
}
int left = maxDepth(root.left) + 1;
int right = maxDepth(root.right) + 1;
return Math.max(left, right);
}
}
Complexity analysis
- Time complexity : we visit each node exactly once, thus the time complexity is \(O(N)\), where \(N\) is the number of nodes.
- Space complexity : in the worst case, the tree is completely unbalanced, e.g. each node has only left child node, the recursion call would occur \(N\) times (the height of the tree), therefore the storage to keep the call stack would be \(O(N)\). But in the best case (the tree is completely balanced), the height of the tree would be \(log(N)\). Therefore, the space complexity in this case would be \(O(log(N))\).
Top-Down
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public int result;
public int maxDepth(TreeNode root) {
result = 0;
dfs(root, 0);
return result;
}
private void dfs(TreeNode root, int depth){
if (root == null){
return;
}
depth++;
result = Math.max(result, depth);
dfs(root.left, depth);
dfs(root.right, depth);
}
}