Given a binary tree, find its maximum depth.
The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.
思路一:递归
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public int maxDepth(TreeNode root) {
if(root==null)
return 0;
else if(root.left==null&&root.right==null)
return 1;
else if(root.left==null)
return maxDepth(root.right)+1;
else if(root.right==null)
return maxDepth(root.left)+1;
return Math.max(maxDepth(root.left),maxDepth(root.right))+1;
}
}
import java.util.*;
public class Solution {
public int maxDepth(TreeNode root) {
if(root==null)
return 0;
return Math.max(maxDepth(root.left),maxDepth(root.right))+1;
}
}
思路二:BFS,效率高,用queue
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
import java.util.*;
public class Solution {
public int maxDepth(TreeNode root) {
if(root==null)
return 0;
Queue<TreeNode> queue=new LinkedList<TreeNode>();
queue.add(root);
int depth=0;//区别于最小depth的时候
while(!queue.isEmpty()){
int size=queue.size();
for(int i=0;i<size;i++){
TreeNode node=queue.poll();
if(node.left!=null)
queue.add(node.left);
if(node.right!=null)
queue.add(node.right);
}
depth++;
}
return depth;
}
}
思路三:DFS,用栈
import java.util.*;
public class Solution {
public int maxDepth(TreeNode root) {
if(root==null)
return 0;
Stack<TreeNode> nodestack=new Stack<TreeNode>();//利用栈的后进先出
Stack<Integer> depthstack=new Stack<>();//对应节点的深度
nodestack.push(root);
depthstack.push(1);
int max=0;
while(!nodestack.isEmpty()){
TreeNode node=nodestack.pop();
int temp=depthstack.pop();
max=Math.max(temp,max);//每个节点,计算当前节点最大深度
if(node.left!=null){//有子节点则加入,深度为当前深度加1;
nodestack.push(node.left);
depthstack.push(temp+1);
}
if(node.right!=null){
nodestack.push(node.right);
depthstack.push(temp+1);
}
}
return max;
}
}
最小深度和最大深度对比
利用递归时:
都可以判断
root null:return 0
else if root.leftnull &&root.rightnull :return 1;
else if root.leftnull: return max/minDepth(root.right)+1;
else if root.right==null: return max/minDepth(root.left)+1;
最后return Math.max/min(max/minDepth(root.left),max/minDepth(root.right))+1;
利用BFS:
最大深度初始depth=0,每层遍历完depth++,最后返回depth
最小深度初始depth=1,遇到左右节点都为空,则返回depth;否则depth++,最后返回depth