树的非递归遍历

1.先序（我只把这些当模板）

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

2.中序

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

3.后序

class Solution {public:    vector<int> postorderTraversal(TreeNode* root) {        vector<int> res;        if(root==NULL)return res;        stack<TreeNode*> ms;        TreeNode *p=root,*pre=NULL;        while(!ms.empty()||p)        {            if(p)            {                ms.push(p);                p=p->left;            }            else            {                p=ms.top();                if(p->right!=NULL&&p->right!=pre)p=p->right;                else                 {                    res.push_back(p->val);                    ms.pop();                    pre=p;                    p=NULL;                }            }        }        return res;    }};

看看大神的代码（以下代码来源于leetcode）

1.先序（看起来很优美）

class Solution {public:vector<int> preorderTraversal(TreeNode *root) {    if (root==NULL) {        return vector<int>();    }    vector<int> result;    stack<TreeNode *> treeStack;    treeStack.push(root);    while (!treeStack.empty()) {        TreeNode *temp = treeStack.top();        result.push_back(temp->val);        treeStack.pop();        if (temp->right!=NULL) {            treeStack.push(temp->right);        }        if (temp->left!=NULL) {            treeStack.push(temp->left);        }    }    return result;}};

2.中序( o(1)空间复杂度）

Morris traversal

public class TreeNode {    int val;    TreeNode left;    TreeNode right;    TreeNode(int x) { val = x; }}public class Solution {    public List<Integer> inorderTraversal(TreeNode root) {        if(root == null) return new ArrayList<Integer>();        List<Integer> res = new ArrayList<Integer>();        TreeNode pre = null;        while(root != null){        if(root.left == null){        res.add(root.val);        root = root.right;        }else{        pre = root.left;        while(pre.right != null && pre.right != root){        pre = pre.right;        }        if(pre.right == null){        pre.right = root;        root = root.left;        }else{        pre.right = null;        res.add(root.val);        root = root.right;        }        }        }        return res;    }}

3.后序（美得不敢直视）

vector<int> postorderTraversal(TreeNode *root) {    vector<int> v;    if (!root) return v;        stack<TreeNode *> s;    s.push(root);        TreeNode *p = NULL;    while(!s.empty()) {        p = s.top();        s.pop();        v.insert(v.begin(), p->val);        if (p->left) s.push(p->left);        if (p->right) s.push(p->right);    }        return v;}