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

PiscesYE 發(fā)布于2019-08-16 13:28 / 2243人閱讀

LeetCode 104 Maximum Depth of Binary Tree
難度：Easy

題目描述：
找到一顆二叉樹的最深深度。
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.

Note: A leaf is a node with no children.

Example:

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

return its depth = 3.

解題思路：
Solution 1：
解二叉樹的題目一般如果難度為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(jù)這一特性，我們自底向上，時(shí)間復(fù)雜度O(n), Space O(1), 但因?yàn)槭沁f歸，所以會占用stack，recursive的寫法如下：

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(gè)Queue，每往下一層都加1，最后拿到最深深度。
時(shí)間：O(n), 空間 O(n)

    public int maxDepth(TreeNode root) {
        //iterative
        if (root == null) return 0;
        Queue 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;
        
    }