104. Maximum Depth of Binary Tree（DFS）

1. Description

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.

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2. Analysis

问题转为求该二叉树的高度，只要计算出该二叉树有几层即可，是一道求二叉树高度的基础题，问题关键是保存遍历过的节点，从而遍历该些节点的子节点。用vector或queue皆可。难度系数：easy

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3. Algorithm achievement

算法的时间复杂度为O(n).

/** * 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;        TreeNode * node = NULL;        int level = 0;        queue<TreeNode*> tmp;        tmp.push(root);        while( !tmp.empty()) {            level ++;            for(int i = 0, size = tmp.size(); i < size; i++) {                node = tmp.front();                tmp.pop();                if(node->left != NULL) {                    tmp.push(node->left);                }                if(node->right != NULL) {                    tmp.push(node->right);                }                       }           }        return level;       }};

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