C++实现LeetCode(104.二叉树的最大深度)

[LeetCode] 104. Maximum Depth of Binary Tree 二叉树的最大深度

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],

3
/ \
9 20
/ \
15 7

return its depth = 3.

求二叉树的最大深度问题用到深度优先搜索 Depth First Search，递归的完美应用，跟求二叉树的最小深度问题原理相同，参见代码如下：

C++ 解法一：

									class Solution {

									public:

									    int maxDepth(TreeNode* root) {

									        if (!root) return 0;

									        return 1 + max(maxDepth(root->left), maxDepth(root->right));

									    }

									};

Java 解法一：

									public class Solution {

									    public int maxDepth(TreeNode root) {

									        return root == null ? 0 : (1 + Math.max(maxDepth(root.left), maxDepth(root.right)));

									    }

									}

我们也可以使用层序遍历二叉树，然后计数总层数，即为二叉树的最大深度，注意 while 循环中的 for 循环的写法有个 trick，一定要将 q.size() 放在初始化里，而不能放在判断停止的条件中，因为q的大小是随时变化的，所以放停止条件中会出错，参见代码如下：

C++ 解法二：

									class Solution {

									public:

									    int maxDepth(TreeNode* root) {

									        if (!root) return 0;

									        int res = 0;

									        queue<TreeNode*> q{{root}};

									        while (!q.empty()) {

									            ++res;

									            for (int i = q.size(); i > 0; --i) {

									                TreeNode *t = q.front(); q.pop();

									                if (t->left) q.push(t->left);

									                if (t->right) q.push(t->right);

									            }

									        }

									        return res;

									    }

									};

Java 解法二：

									public class Solution {

									    public int maxDepth(TreeNode root) {

									        if (root == null) return 0;

									        int res = 0;

									        Queue<TreeNode> q = new LinkedList<>();

									        q.offer(root);

									        while (!q.isEmpty()) {

									            ++res;

									            for (int i = q.size(); i > 0; --i) {

									                TreeNode t = q.poll();

									                if (t.left != null) q.offer(t.left);

									                if (t.right != null) q.offer(t.right);

									            }

									        }

									        return res;

									    }

									}