LeetCode104 maximum-depth-of-binary-tree

题目

Given the root of a binary tree, return its maximum depth.

A binary tree's maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

 Example 1: 
Input: root = [3,9,20,null,null,15,7]
Output: 3

 Example 2: 
Input: root = [1,null,2]
Output: 2

 Example 3: 
Input: root = []
Output: 0

 Example 4: 
Input: root = [0]
Output: 1

 Constraints: 
 The number of nodes in the tree is in the range [0, 10⁴]. 
 -100 <= Node.val <= 100 

方法

递归法

max(left,right)+1

• 时间复杂度：O(n)，n是二叉树节点的个数
• 空间复杂度：O(h)，h是二叉树的高度

Java版本：

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

Js版本：

var maxDepth = function(root) {
    if(root==null) return 0;
    return Math.max(maxDepth(root.left),maxDepth(root.right))+1;
};

迭代法

类似层序遍历二叉树，只是每次一行的节点遍历完之后才增加层数

• 时间复杂度：O(n)，n是二叉树节点的个数
• 空间复杂度：此方法空间的消耗取决于队列存储的元素数量，其在最坏情况下会达到 O(n)

Java版本：

class Solution {
    public int maxDepth(TreeNode root) {
        if(root==null){
            return 0;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        int h = 0;
        while(!queue.isEmpty()){
            h++;
            int len = queue.size();
            for(int i=0;i<len;i++){
                TreeNode node = queue.poll();
                if(node.left!=null){
                    queue.offer(node.left);
                }
                if(node.right!=null){
                    queue.offer(node.right);
                }
            }
        }
        return h;
    }
}

Js版本：

var maxDepth = function(root) {
    if(root==null)return 0;
    let max = 0;
    const queue = [];
    queue.push(root);
    while (queue.length){
        max++;
        let size = queue.length;
        for(let i=0;i<size;i++){
            let node = queue.shift();
            if(node.left!==null) queue.push(node.left);
            if(node.right!==null) queue.push(node.right);
        }
    }
    return max;
};