Minimum Depth of Binary Tree
Find the shortest path length from root to any of the leaves. A simple BFS problem.
Note that using DFS in this problem is vulnerable to StackOverflowError
.
We are using a LinkedList
as queue and using two pointer to mark end of a level.
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class MinimumDepthOfBinaryTree {
public int minDepth(TreeNode root) {
if (root == null) return 0;
LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
queue.add(root);
TreeNode last = root;
TreeNode nlast = null;
int depth = 1;
while (!queue.isEmpty()) {
TreeNode node = queue.pollFirst();
if (node.left != null) {
queue.addLast(node.left);
nlast = node.left;
}
if (node.right != null) {
queue.addLast(node.right);
nlast = node.right;
}
if (node.left == null && node.right == null) return depth;
if (node == last) {
last = nlast;
depth++;
}
}
return depth;
}
}
The code below uses DFS. It also get an AC on LeetCode.
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class MinimumDepthOfBinaryTree {
public int minDepth(TreeNode root) {
if (root == null) return 0;
if (root.left != null && root.right != null) {
return Math.min(minDepth(root.left), minDepth(root.right)) + 1;
} else if (root.left != null) {
return minDepth(root.left) + 1;
} else {
return minDepth(root.right) + 1;
}
}
}