轉載：【刷題】二叉樹非遞迴遍歷

stack<Node*> st;


void preOrder(Node* root) {
    Node *cur = root;
    while (cur || !st.empty()) {
        while (cur) {
            //訪問、入棧、轉向左孩子
            pre.push_back(cur->data);
            st.push(cur);
            cur = cur->lchild;

        }

        
 
if (!st.empty()) {
            //出棧、轉向右孩子
            cur = st.top();
            st.pop();
            cur = cur->rchild;
        }
    }
}

void inOrder(Node* root) {
    Node *cur = root;
    while (cur || !st.empty()) {
        while (cur) {
            //入棧、轉向左孩子
            st.push(cur);
            cur 
 
= cur->rchild;
        }

        if (!st.empty()) {
            //出棧、訪問、轉向右孩子
            cur = st.top();
            st.pop();
            in.push_back(cur->data);
            cur = cur->rchild;
        }
    }
}

void postOrder(Node* root) {
    Node *cur = root, *pre = NULL;//pre指向上一個被訪問過的節點
    
 
while (cur || !st.empty()) {
        while (cur) {
            //入棧、轉向左孩子
            st.push(cur);
            cur = cur->lchild;
        }

        if (!st.empty()) {
            //有條件的出棧，
            cur = st.top();
            // 如果已經訪問了右子樹，則可以訪問根節點；否則轉向，先去訪問右子樹

            if (cur->rchild == NULL || cur->rchild == pre) {
                st.pop();
                post.push_back(cur->data);
                pre = cur;
                cur = NULL;// 表示不需要轉向，繼續彈棧

            }
            else {
                //轉向
                cur = cur->rchild;
            }
        }
    }
}