[LintCode] 155 Minimum Depth of Binary Tree 解题报告

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.


Example
Given a binary tree as follow:

  1
 / \
2   3
    / \
  4   5
The minimum depth is 2.


思路
解法一
BFS
一层一层往queue里放node。如果发现某一个node左右都是null，说明找最短路径到这个点就停了。当前的level就是要求的minimum depth。


Code

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: An integer.
     */
    public int minDepth(TreeNode root) {
        // write your code here
        if (root == null) {
            return 0;
        }
        
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        int currLevel = 0;
        
        while (!queue.isEmpty()) {
            int size = queue.size();
            currLevel++;
            for (int i = 0 ; i < size; i++) {
                TreeNode node = queue.poll();
                if (node.left != null) {
                    queue.offer(node.left);
                }
                if (node.right != null) {
                    queue.offer(node.right);
                }
                if (node.left == null && node.right == null) {
                    return currLevel;
                }
            }
        }
        
        return currLevel;
    }
}