Maximum Depth of Binary Tree

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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.

分析：
同上题

代码：

#include <iostream>#include <vector>#include <stack>#include <algorithm>#include <queue>#define Elementype intusing namespace std;int index = 0;typedef struct TreeNode //树结点{    Elementype  val;    TreeNode *left;    TreeNode *right;    TreeNode(Elementype x) : val(x), left(nullptr), right(nullptr) {}} *Tree;//使用先序遍历创建创建二叉树void MakeBinaryTree(Tree &T, Elementype value[]){    Elementype  c = value[index++];    if (c == '#')        T = nullptr;    else    {        T = new TreeNode(c);        MakeBinaryTree(T->left, value);        MakeBinaryTree(T->right, value);    }}//递归int maxDepth(Tree root){    if (root == nullptr)        return 0;    return max(maxDepth(root->left),maxDepth(root->right)) + 1;}//迭代，广度优先遍历int maxDepthBFS(Tree root){    if (!root)        return 0;    int depth = 1;    queue<Tree> que;    queue<Tree> que2;    que.push(root);    while (!que.empty())    {        Tree p = que.front();        que.pop();        if (p->left)            que2.push(p->left);        if (p->right)            que2.push(p->right);        if (que.empty() && !que2.empty())        {            swap(que, que2);            depth++;        }    }    return depth;}//先序遍历void preOrderTraversal(Tree root){    stack<Tree> sk;    Tree p = root;    if (p)        sk.push(p);    while (!sk.empty())    {        p = sk.top();        cout << p->val<<" ";        sk.pop();        if (p->right)            sk.push(p->right);        if (p->left)            sk.push(p->left);    }}int main(){    Tree root = nullptr;        //根结点    Elementype data[11] = { 3, 9, '#', '#', 20, 15, '#', '#', 7, '#', '#' };    MakeBinaryTree(root, data);    cout << "先序遍历： ";    preOrderTraversal(root);    cout << endl;    cout << "最大深度，递归：" << maxDepth(root) << endl;    cout << "最大深度，迭代：" << maxDepthBFS(root) << endl;    system("pause");    return 0;}

测试：

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