算法——Week12

104. Maximum Depth of Binary Tree
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.

解题思路
使用递归。根节点的深度 = max（左子树深度，右子树深度）。

代码如下：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int maxDepth(TreeNode* root) {
        if(root == NULL) {
            return 0;
        }
        //这么写会报超时，用max函数则不会
        /*return maxDepth(root->left) > maxDepth(root->right) ? maxDepth(root->left) + 1 : maxDepth(root->right) + 1;
        */
       return max(1 + maxDepth(root -> left), 1 + maxDepth(root -> right));
    }
};

上述递归解法是基于深度优先搜索，用广度优先搜索也可以解题。代码如下：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int maxDepth(TreeNode* root) {
        if(root == NULL)
            return 0;
        queue<TreeNode*> q;
        q.push(root);
        int result = 0;
        while(q.size() != 0) {
            result++;
            int size = q.size();
            for(int i = 0; i < size; i++) {
                TreeNode *temp = q.front();
                q.pop();
                if(temp->left != NULL) {
                    q.push(temp->left);
                }
                if(temp->right != NULL) {
                    q.push(temp->right);
                }
            }
        }
        return result;
    }
};

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