LeetCode 104 Maximum Depth of Binary Tree 二叉樹最大深度

LeetCode 104 Maximum Depth of Binary Tree
難度:Easynode

題目描述:
找到一顆二叉樹的最深深度。
Given a binary tree, find its maximum depth.code

The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.遞歸

Note: A leaf is a node with no children.ip

Example:leetcode

Given binary tree [3,9,20,null,null,15,7],get

3
   / \
  9  20
    /  \
   15   7

return its depth = 3.it

解題思路:
Solution 1:
解二叉樹的題目通常若是難度爲easy,則要求iterative和recursive都會寫,二叉樹的最深深度是左子樹和右子樹的Max深度,根據這一特性,咱們自底向上,時間複雜度O(n), Space O(1), 但由於是遞歸,因此會佔用stack,recursive的寫法以下:io

public int maxDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        //get the left and right depth
        int leftDep = maxDepth(root.left);
        int rightDep = maxDepth(root.right);
        return 1 + Math.max(leftDep, rightDep);
    }

解題思路:
Solution 2:
使用BFS方法iterative使用一個Queue,每往下一層都加1,最後拿到最深深度。
時間:O(n), 空間 O(n)List

public int maxDepth(TreeNode root) {
        //iterative
        if (root == null) return 0;
        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);
        int dep = 0;
        while(!q.isEmpty()) {
            dep++;
            int size = q.size();
            for(int i = 0; i < size; i++) {
                TreeNode now = q.poll();
                if (now.left != null) q.offer(now.left);
                if (now.right != null) q.offer(now.right);
            }

        }
        return dep;
        
    }
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