[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.

---

/* 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;
    }
};

---

/** 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;
    }
};