Binary Tree Inorder Traversal

Given a binary tree, return the inorder traversal of its nodes' values. 中序遍历

Input: [1,null,2,3]
1

2
/
3

Output: [1,3,2]

解法

递归

struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 };

class Solution {
public:
      vector<int> inorderTraversal(TreeNode *root) {
        vector<int> res;
        inorder(root, res);
      return res;
    }
    void inorder(TreeNode *root, vector<int> &res) {
        if (!root) return;
        if (root->left) inorder(root->left, res);
        res.push_back(root->val);
        if (root->right) inorder(root->right, res);
    }
};

非递归

用栈来代替递归，简历二叉树指针栈，储存所有的遍历的二叉树节点。每次储存后，中序遍历移动指针到当前左子节点，直到遍历到最左侧叶子节点，对于最左侧叶子节点，取当前节点值，并指针移动到当前节点右子节点，栈中删除当前节点。循环直到栈为空并且指针指向空节点。（有左边，去左边，没左边，打印当前，去最右边。）
class Solution {
public:

    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> res;
        stack<TreeNode* > s;
        TreeNode* p = root;
        while(p!= NULL || !s.empty()){
            if (p!=NULL){
                s.push(p);
                p = p->left;
            }else{
                res.push_back(s.top()->val);
                p = s.top()->right;
                s.pop();
            }
        }
        return res;
    }
};

Maximum Depth of Binary Tree

class Solution {
public:
    int maxDepth(TreeNode* root) {
        if (root == NULL) return 0;
        return max(maxDepth(root->left), maxDepth(root->right))+1;

    }
};

Minimum Depth of Binary Tree

class Solution {
public:
    int minDepth(TreeNode* root) {
        if(root == NULL) return 0;
        if (root->left == NULL && (root->right == NULL) ) return 1; 
        if (root->left == NULL) return minDepth(root->right)+1;
        if (root->right == NULL) return minDepth(root->left)+1;
        return min(minDepth(root->left), minDepth(root->right))+1;

    }
};