栈和队列遍历二叉树

栈深度优先遍历

算法思想：

1、得到根结点后，若左孩子结点不空，不断地将当前结点的左孩子则入栈，第一步完成后，栈顶结点是叶结点
2、栈不空时，将栈顶元素弹出并保存到返回的答案数组，并查看当前结点是否有右孩子，有则入栈，然后重复步骤1。没有则接着执行步骤2，直到栈空。

class Solution {
public:
    vector<int> res; 
    stack<TreeNode*> s;
    vector<int> inorderTraversal(TreeNode* root){
        if(root == nullptr)  return res;
        TreeNode* cur = root;  
        s.push(cur);
        //首先将所有的左孩子压栈
        while(cur->left != nullptr){
            s.push(cur->left);
            cur = cur->left;
        }
        //退出循环后cur指向叶结点，准备回溯
        while(!s.empty()){
            cur = s.top();
            res.push_back(cur->val);
            s.pop();
            if(cur->right != nullptr){
                s.push(cur->right);
                cur = cur->right;
                while(cur->left != nullptr){
                    s.push(cur->left);
                    cur = cur->left;
                }
            }
        }
        return res;
    }
};

递归解法

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

队列广度优先遍历

算法思想：

1. 根结点入队后，然后将所有不为空的子结点入队
2. 每次循环进行操作时，都要同时处理一层的子结点，即将此时队列中所有结点的子结点全部入队，同时将当前处理完成的父结点出队

vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> res;
        if(root == nullptr){
            return res;
        }
        queue<TreeNode*> q;
        q.push(root);
        while(!q.empty()){
            vector<int> temp;
            int count = q.size();
            for(int i=1; i<=count; i++){
                TreeNode* cur = q.front();
                if(cur->left != nullptr){
                    q.push(cur->left);
                }
                if(cur->right != nullptr){
                    q.push(cur->right);
                }
                temp.push_back(cur->val);
                q.pop();
            }
            res.push_back(temp);
        }
        return res;
    }

“之”字型打印结点

双端队列 + 层序遍历

1. 遍历奇数层时：从队首出队，其子结点 先左后右从队尾入队
2. 遍历偶数层时：从队尾出队，其子结点 先右后左从队头入队

vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> res;
        if(root == nullptr){
            return res;
        }
        deque<TreeNode*> dq;
        dq.push_back(root);
        bool isOdd = true;
        while(!dq.empty()){
            vector<int> layer;
            int len = dq.size();
            // 奇数层
            if(isOdd){
                while(len-- > 0){
                    TreeNode* node = dq.front();
                    layer.push_back(node->val);
                    dq.pop_front();
                    if(node->left != nullptr){
                        dq.push_back(node->left);
                    }
                    if(node->right != nullptr){
                        dq.push_back(node->right);
                    }
                }
            }else{
                while(len-- > 0){
                    TreeNode* node = dq.back();
                    layer.push_back(node->val);
                    dq.pop_back();
                    if(node->right != nullptr){
                        dq.push_front(node->right);
                    }
                    if(node->left != nullptr){
                        dq.push_front(node->left);
                    }
                }
            }
            res.push_back(layer);
            isOdd = !isOdd;
        }
        return res;
    }

N叉树的前序遍历

结点结构

class Node {
public:
    int val;
    vector<Node*> children;

    Node() {}

    Node(int _val) {
        val = _val;
    }

    Node(int _val, vector<Node*> _children) {
        val = _val;
        children = _children;
    }
};

递归版1

class Solution {
public:
    vector<int> res;

    void traverse(Node* root){
        if(root == nullptr){
            return ;
        }
        res.push_back(root->val);
        for(Node* node : root->children){
            preorder(node);
        }
    }

    vector<int> preorder(Node* root) {
        tranverse(root);
        return res;
    }
};

递归版2

class Solution {
public:
    vector<int> res;

    vector<int> preorder(Node* root) {
        if(root == nullptr){
            return vector<int>();
        }
        res.push_back(root->val);
        for(Node* node : root->children){
            preorder(node);
        }
        return res;
    }
};

迭代版

class Solution {
public:
    vector<int> res;

    vector<int> preorder(Node* root) {
        if(root == nullptr){
            return res;
        }
        stack<Node*> s;
        s.push(root);
        while(!s.empty()){
            Node* cur = s.top();
            s.pop();
            res.push_back(cur->val);
            for(int i = cur->children.size() - 1; i >= 0; i--){
                s.push(cur->children[i]);
            }
        }
        return res;
    }
};