题解 | 二叉树的最小深度

import java.util.*;

/*
 * public class TreeNode {
 *   int val = 0;
 *   TreeNode left = null;
 *   TreeNode right = null;
 *   public TreeNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

/**
 * NC234 二叉树的最小深度
 * @author d3y1
 */
public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param root TreeNode类 
     * @return int整型
     */
    public int run (TreeNode root) {
        // return solution1(root);
        return solution11(root);
        // return solution2(root);
    }

    /**
     * 递归: 后序遍历
     * @param root
     * @return
     */
    private int solution1(TreeNode root){
        return postOrder(root);
    }

    /**
     * 后序遍历
     * @param root
     * @return
     */
    private int postOrder(TreeNode root){
        if(root == null){
            return 0;
        }

        int leftDep = postOrder(root.left);
        int rightDep = postOrder(root.right);

        int curDep;
        if(leftDep>0 && rightDep>0){
            curDep = Math.min(leftDep, rightDep)+1;
        }else if(leftDep>0 && rightDep==0){
            curDep = leftDep+1;
        }else if(leftDep==0 && rightDep>0){
            curDep = rightDep+1;
        }else{
            curDep = 1;
        }

        return curDep;
    }

    private int result = Integer.MAX_VALUE;

    /**
     * 递归: 后序遍历
     * @param root
     * @return
     */
    private int solution11(TreeNode root){
        if(root == null){
            return 0;
        }

        postorder(root, 1);

        return result;
    }

    /**
     * 后序遍历
     * @param root
     * @param depth
     */
    private void postorder(TreeNode root, int depth){
        if(root == null){
            return;
        }

        if(depth < result){
            postorder(root.left, depth+1);
            postorder(root.right, depth+1);
            // 当前节点 叶子节点
            if(root.left==null && root.right==null){
                result = depth;
            }
        }
    }

    /**
     * 迭代: 层序遍历
     * @param root
     * @return
     */
    private int solution2(TreeNode root){
        return levelOrder(root);
    }

    /**
     * 层序遍历
     * @param root
     * @return
     */
    private int levelOrder(TreeNode root){
        if(root == null){
            return 0;
        }

        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);

        int dep = 0;
        int size;
        TreeNode node;
        while(!queue.isEmpty()){
            size = queue.size();
            dep++;
            while(size-- > 0){
                node = queue.poll();
                // 第一个 叶子节点
                if(node.left==null && node.right==null){
                    return dep;
                }else{
                    if(node.left != null){
                        queue.offer(node.left);
                    }
                    if(node.right != null){
                        queue.offer(node.right);
                    }
                }
            }
        }

        return dep;
    }
}