二叉树前序、中序和后序的非递归遍历

把经典问题：二叉树的非递归遍历理了理，顺便做个笔记。
核心思路就是节点的压栈和出栈。

#include <iostream>#include<stack>#include<vector>using namespace std;struct TreeNode {    int val;    TreeNode* left;    TreeNode* right;    TreeNode(int x) : val(x) { left = right = nullptr; }    TreeNode(int x, TreeNode* theLeft, TreeNode* theRight)        : val(x), left(theLeft), right(theRight) { }};class Solution {    TreeNode* root;    void print(vector<int> &result, const char* str) {        vector<int>::iterator it = result.begin();        cout << str;        for (; it != result.end(); ++it)            cout << *it << " ";        cout << endl;    }public:    Solution(TreeNode* theRoot) : root(theRoot) { }    TreeNode* getRoot() { return root; }    //递归遍历    void prePrint_D(TreeNode* root) {        if (root) {            cout << root->val << " ";            prePrint_D(root->left);            prePrint_D(root->right);        }    }    void midPrint_D(TreeNode* root) {        if (root) {            midPrint_D(root->left);            cout << root->val << " ";            midPrint_D(root->right);        }    }    void postPrint_D(TreeNode* root) {        if (root) {            postPrint_D(root->left);            postPrint_D(root->right);            cout << root->val << " ";        }    }    //非递归遍历    void prePrint() {        if (!root)            return;        stack<TreeNode*> nodeStack;        vector<int> result;        TreeNode* cNode = root;        while (cNode || !nodeStack.empty()) {            result.push_back(cNode->val);            nodeStack.push(cNode);            cNode = cNode->left;            while (!cNode && !nodeStack.empty()) {                cNode = nodeStack.top();                nodeStack.pop();                cNode = cNode->right;            }        }        print(result, "preOrder: ");    }    void midPrint() {        if (!root)            return;        stack<TreeNode*> nodeStack;        vector<int> result;        TreeNode* cNode = root;        while (cNode || !nodeStack.empty()) {            if (cNode->left) {                nodeStack.push(cNode);                cNode = cNode->left;            }            else {                result.push_back(cNode->val);                cNode = cNode->right;                while (!cNode && !nodeStack.empty()) {                    cNode = nodeStack.top();                    nodeStack.pop();                    result.push_back(cNode->val);                    cNode = cNode->right;                }            }        }        print(result, "midOrder: ");    }    void postPrint() {        if (!root)            return;        stack<TreeNode*> nodeStack;        vector<int> result;        TreeNode* cNode = root;        nodeStack.push(cNode);        while (!nodeStack.empty()) {            cNode = nodeStack.top();            if ((!cNode->left && !cNode->right) ||   //左右子树为空                (!result.empty() && cNode->right &&                     cNode->right->val == result[result.size() - 1])    //有右子树，右子树刚遍历完                || (!result.empty() && cNode->left &&                     cNode->left->val == result[result.size() - 1])) {    //没有右子树，但左子树刚遍历完                //此时cNode节点的子树已全部遍历完，输出该节点，并从栈里删除                result.push_back(cNode->val);                nodeStack.pop();            }            else {    //将左右子树压入栈，由于遍历顺序是先左后右，所以入栈顺序是先右后左                if (cNode->right) nodeStack.push(cNode->right);                if (cNode->left) nodeStack.push(cNode->left);            }        }        print(result, "postOrder: ");    }};int main(){    TreeNode* p1 = new TreeNode(1),        *p2 = p1->left = new TreeNode(2),        *p3 = p1->right = new TreeNode(3),        *p4 = p2->left = new TreeNode(4),        *p5 = p2->right = new TreeNode(5),        *p6 = p3->left = new TreeNode(6),        *p7 = p3->right = new TreeNode(7),        *p8 = p4->left = new TreeNode(8),        *p9 = p4->right = new TreeNode(9),        *p10 = p5->left = new TreeNode(10),        *p11 = p5->right = new TreeNode(11),        *root = p1;    Solution S(root);    cout << "递归：" << endl;    S.prePrint();    S.midPrint();    S.postPrint();    cout << "递归：" << endl;    cout << "preOrder: ";    S.prePrint_D(S.getRoot());    cout << endl;    cout << "midOrder: ";    S.midPrint_D(S.getRoot());    cout << endl;    cout << "postOrder: ";    S.postPrint_D(S.getRoot());    cout << endl;    //...释放二叉树内存代码段...    return 0;}

输出：

递归：preOrder: 1 2 4 8 9 5 10 11 3 6 7 midOrder: 8 4 9 2 10 5 11 1 6 3 7 postOrder: 8 9 4 10 11 5 2 6 7 3 1 递归：preOrder: 1 2 4 8 9 5 10 11 3 6 7 midOrder: 8 4 9 2 10 5 11 1 6 3 7 postOrder: 8 9 4 10 11 5 2 6 7 3 1