111. Minimum Depth of Binary Tre

Description

Given a binary tree, find its minimum depth.

The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.

Note: A leaf is a node with no children.

Example:

Given binary tree [3,9,20,null,null,15,7],

return its minimum depth = 2.

Solution

DFS, time O(n), space O(h)

题目有坑啊，一定注意，是从root到left的minDepth！而非root到null的！如果root.right == null，minDepth不是1，而是应该计算root到leftTree的leaf才对。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int minDepth(TreeNode root) {
        if (root == null) {
            return 0;
        } else if (root.left == null) {
            return 1 + minDepth(root.right);
        } else if (root.right == null) {
            return 1 + minDepth(root.left);
        } else {
            return 1 + Math.min(minDepth(root.left), minDepth(root.right));
        }
    }
}

BFS, time O(n), space O(n)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int minDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        int depth = 1;
        
        while (!queue.isEmpty()) {
            int size = queue.size();
            
            while (size-- > 0) {
                TreeNode curr = queue.poll();
                if (curr.left == null && curr.right == null) {
                    return depth;
                }
                
                if (curr.left != null) {
                    queue.offer(curr.left);
                }
                
                if (curr.right != null) {
                    queue.offer(curr.right);
                }
            }
            
            ++depth;
        }
        
        return -1;
    }
}