剑指offer---38.二叉树的深度

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本文链接： https://blog.csdn.net/weixin_44406146/article/details/102694418

题目描述

输入一棵二叉树，求该树的深度。从根结点到叶结点依次经过的结点（含根、叶结点）形成树的一条路径，最长路径的长度为树的深度。

解法一：递归

/**
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

    public TreeNode(int val) {
        this.val = val;

    }

}
*/
public class Solution {
    public int TreeDepth(TreeNode root) {
        if(root==null) return 0;
        int left = TreeDepth(root.left);
        int right = TreeDepth(root.right);
        return (left>right?left:right)+1;
    }
}

解法二：非递归

思路： 借助层次遍历

import java.util.*;
/**
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

    public TreeNode(int val) {
        this.val = val;

    }
}
*/
public class Solution {
    public int TreeDepth(TreeNode root) {
        if(root==null) return 0;
        ArrayList<TreeNode> list = new ArrayList<>();
        int count = 0;
        int low=0,high=0;
        int tmphigh = high;
        list.add(root);
        while(low<=high){
            TreeNode node = list.get(low);
            if(node.left!=null){
                list.add(node.left);
                high++;
            } 
            if(node.right!=null){
                list.add(node.right);
                high++;
            } 
            low++;
            if(low>tmphigh){
                count++;
                tmphigh = high;
            }
        }
        return count;
    }
}

解法三：非递归用队列

思路： 可节省空间复杂度

/**
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

    public TreeNode(int val) {
        this.val = val;

    }

}
*/

import java.util.Queue;
import java.util.LinkedList;
 
public class Solution {
    public int TreeDepth(TreeNode pRoot) {
        if(pRoot == null){
            return 0;
        }
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.add(pRoot);
        int depth = 0, count = 0, nextCount = 1;
        while(queue.size()!=0){
            TreeNode top = queue.poll();
            count++;
            if(top.left != null){
                queue.add(top.left);
            }
            if(top.right != null){
                queue.add(top.right);
            }
            if(count == nextCount){
                nextCount = queue.size();
                count = 0;
                depth++;
            }
        }
        return depth;
    }
}