[基础算法]二叉树的非递归遍历

/*
    这个程序包含了二叉树的前序遍历和后序遍历的非递归形式， 二叉树的非递归形式有很多种， 这里是我认为比较好记的方式。
*/
#include <iostream>
#include <stack>

using namespace std;

class TreeNode{
public:
    TreeNode(int v){val = v;};
    int val;
    TreeNode* left;
    TreeNode* right;
};

void prevOrder(TreeNode* root){
    TreeNode* curr;
    stack<TreeNode*> s;
    curr = root;
    while(curr || !s.empty()){ //前面有路或者还有退路
            //往左走到头
        while(curr){ 
            cout<<curr->val<<endl;
            s.push(curr);
            curr = curr->left;
        }
            //前面没路了，退回来并试着往右
        if(!s.empty()){ 
            curr = s.top();
            s.pop();
            curr = curr->right;
        }
    }
}


/*
    这个程序改编自多叉树的后序遍历方法。
*/
void postOrder(TreeNode* root){
    stack<TreeNode*> s;
    TreeNode* curr;
    TreeNode* lastPopedNode=NULL;
    s.push(root);
    while(!s.empty()){
        curr = s.top();
        TreeNode* lastChild = NULL;
        if(curr->left) lastChild = curr->left;
        if(curr->right) lastChild = curr->right;

        if(lastChild && lastPopedNode!=lastChild){
            if(curr->right) s.push(curr->right);
            if(curr->left) s.push(curr->left);
        }else{
            cout<<curr->val<<endl;
            s.pop();
            lastPopedNode = curr;
        }
    }
}

//该函数用于建立二叉树
TreeNode* constructTree(int* A, int l ,int r){
    if(l>r) return NULL;
    int mid = (l+r)/2;
    TreeNode* root = new TreeNode(A[mid]);
    root->left  = constructTree(A, l, mid-1);
    root->right = constructTree(A, mid+1, r);
}

int main(){
    int A[] = {1,2,3,4};
    int n = sizeof(A)/sizeof(int);
    TreeNode* root = constructTree(A, 0, n-1);
    postOrder(root);
    return 0;
}