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

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.

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/* 解法一：深度优先遍历（Depth-first-search） * 简洁的递归解法，选择深度较大的子树的长度+1 * 即为整体二叉树的最大深度，依次递归调用即可 *  * 时间复杂度：O(n);空间复杂度：O(lgn) * 解法一实现版本： * 1.递归实现 * 2.递归实现--一行代码（好玩而已） * 3.迭代实现--后序遍历 *//** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */// 1.递归实现class Solution {public:    int maxDepth(TreeNode* root) {        if(root == NULL)            return 0;        return max(maxDepth(root->left), maxDepth(root->right)) + 1;    }};// 2.递归实现--一行代码class Solution {public:    int maxDepth(TreeNode* root) {        return root == NULL ? 0 : max(maxDepth(root->left), maxDepth(root->right)) + 1;    }// 3.迭代实现--后序遍历class Solution {public:    int maxDepth(TreeNode* root) {        // 后序遍历存储结点的临时栈 其最大长度就是二叉树的最大深度        int result = 0;        if(!root)            return result;        stack<TreeNode*> s;        // pCur:当前访问节点，pLastVisit:上次访问节点        TreeNode* pCur = root;        TreeNode* pLastVisit = NULL;        // 走到最左端        while(pCur){            s.push(pCur);            pCur = pCur->left;        }        while(!s.empty()){            result = max(result, int(s.size()));            pCur = s.top();            s.pop();            // 一个根节点被访问的前提是：无右子树或右子树已被访问过             if(pCur->right == NULL || pCur->right == pLastVisit){                pLastVisit = pCur;            }else{                // 根节点再次入栈                s.push(pCur);                //进入右子树，且可肯定右子树一定不为空                pCur = pCur->right;                while(pCur){                    s.push(pCur);                    pCur = pCur->left;                }            }        }        return result;    }};

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/** 解法二：宽度优先遍历（Breadth-first-search） * 其实就是二叉树的层序遍历，有几层，最大深度就是多少 *//** * 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;        int result = 0;        queue<TreeNode*> q;        q.push(root);        while(!q.empty())        {            ++ result;            for(int i = 0, n = q.size(); i < n; ++ i)            {                TreeNode *p = q.front();                q.pop();                if(p->left != NULL)                    q.push(p -> left);                if(p->right != NULL)                    q.push(p -> right);            }        }        return result;    }};