0、二叉树最大深度
原题目：Given a binary tree, find its minimum depth.The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.
方法：求最大深度的时候，只需要比较左右子树的深度，取较大者+1就行了
C++代码：
class Solution
{
public:
int minDepth(TreeNode *root){
if(root==Null) return ;
int l=minDepth(root->left);
int r=minDepth(root->right);
if(l==||r==)
return +l+r;
return +min(l,r);
}
};1、二叉树最小深度
原题目：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.
方法：求最小深度的时候，需要区分双子树与单子树，双子树时，深度较小者+1，单子树时（即左右子树有一颗为空时）为深度较大者+1
C++代码：
class Solution{
public:
int maxDepth(TreeNode *root){
if (root==NULL)
{
return ;
}
int l=maxDepth(root->left);
int r=maxDepth(root->right);
return +max(l,r);
}
};2、Given a binary tree, return the postorder traversal of its nodes' values.
For example:
Given binary tree{1,#,2,3},
1
\
2
/
3
return[3,2,1].
Note: Recursive solution is trivial, could you do it iteratively?
方法：后序遍历，使用vector栈来做，先将父节点入栈，再将右孩子入栈，左孩子入栈。那么返回时就能可以倒序输出后序遍历值。
class Solution{
public:
void postOrder(TreeNode *root,vector<int>&vec){
if (root!=NULL)
{
postOrder(root->left,vec);
postOrder(root->right,vec);
vec.push_back(root->val);
}
}
vector<int>postorderTraversal(TreeNode *root){
vector<int> vec;
postOrder(root,vec);
return vec;
}
};