【leetcode】104. Maximum Depth of Binary Tree【java】三种实现方法：递归、BFS、DFS

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.

/**

 * Definition for a binary tree node.

 * public class TreeNode {

 *     int val;

 *     TreeNode left;

 *     TreeNode right;

 *     TreeNode(int x) { val = x; }

 * }

 */
//方法一：递归
public class Solution {
public int maxDepth(TreeNode root) {
if (root == null){
return 0;
}
return 1 + Math.max(maxDepth(root.left), maxDepth(root.right));
}
}

//方法二 BFS 使用队列 这种方法相对快一些，也好一些
public class Solution {
public int maxDepth(TreeNode root) {
if (root == null){
return 0;
}
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.add(root);
int count = 0;
while (!queue.isEmpty()){
int size = queue.size();
for (int i = size; i > 0; i--){
TreeNode node = queue.poll();
if (node.left != null){
queue.add(node.left);
}
if (node.right != null){
queue.add(node.right);
}
}
count++;
}
return count;
}
}

//方法3 DFS 使用栈
public class Solution {
public int maxDepth(TreeNode root) {
if (root == null){
return 0;
}
Stack<TreeNode> stack = new Stack<>();
Stack<Integer> value = new Stack<>();
int max = 0;
stack.push (root);
value.push(1);
while (!stack.isEmpty()){
TreeNode node = stack.pop();
int temp = value.pop();
max = Math.max(temp, max);
if (node.left != null){
stack.push(node.left);
value.push(temp + 1);
}
if (node.right != null){
stack.push(node.right);
value.push(temp + 1);
}
}
return max;
}
}