二叉树（LeetCode） C++相关知识代码 系列1

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 0;
        int l=minDepth(root->left);
        int r=minDepth(root->right);
        if(l==0||r==0)
            return 1+l+r;
        return 1+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 0;
        }
        int l=maxDepth(root->left);
        int r=maxDepth(root->right);
        return 1+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;
        }

};