给定一个二叉树,找出其最大深度。
二叉树的深度为根节点到最远叶子节点的最长路径上的节点数。
说明: 叶子节点是指没有子节点的节点。
示例:
给定二叉树 [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
返回它的最大深度 3 。
思路1:递归
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int maxDepth(TreeNode root) {
if(root == null) return 0;//退出条件
int leftMaxDepth = maxDepth(root.left);
int rightMaxDepth = maxDepth(root.right);
return Math.max(leftMaxDepth,rightMaxDepth)+1;
}
}
思路2:BFS 二叉树层次遍历
class Solution {
public int maxDepth(TreeNode root) {
if(root == null) return 0;
LinkedList<TreeNode> queue=new LinkedList<>();
queue.add(root);
int maxDepth=0;
while(!queue.isEmpty()){
maxDepth++;
int levelSize=queue.size();
for(int i=0;i<levelSize;i++){
TreeNode current = queue.pollFirst();
if(current.left != null)
queue.add(current.left);
if(current.right != null)
queue.add(current.right);
}
}
return maxDepth;
}
}
思路3:DFS
import javafx.util.Pair;
class Solution {
public int maxDepth(TreeNode root) {
if(root == null) return 0;
Stack<Pair<TreeNode,Integer>> stack =new Stack<>();
stack.push(new Pair<>(root,1));
int max =1;
while(!stack.isEmpty()){
Pair<TreeNode,Integer> current = stack.pop();
Integer currentDepth=current.getValue();
TreeNode currentNode=current.getKey();
if(currentNode!= null){
max=Math.max(currentDepth,max);
stack.push(new Pair<>(currentNode.left,currentDepth+1));
stack.push(new Pair<>(currentNode.right,currentDepth+1));
}
}
return max;
}
}