12. Binary Tree Postorder Traversal && Binary Tree Preorder Traversal

详见：剑指 Offer 题目汇总索引：第6题

Binary Tree Postorder Traversal

          

Given a binary tree, return the postorder traversal of its nodes' values.

For example: Given binary tree {1,#,2,3},

   1
    
     2
    /
   3

return [3,2,1].

Note: Recursive solution is trivial, could you do it iteratively?

注：后序遍历是较麻烦的一个，不可大意。关键两点： 1.要走到 p->left | p->right ==0, 2.每次出栈出两个结点。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> postorderTraversal(TreeNode *root) {
        vector<int> answer;
        if(root == NULL) return answer;
        stack<TreeNode*> st;
        st.push(root);
        TreeNode *p = root;
        while(p->right || p->left) {
            while(p->left) { st.push(p->left); p = p->left;}
            if(p->right) { st.push(p->right); p = p->right;}
        }
        while(!st.empty()) {
            TreeNode *q = st.top(); st.pop();
            answer.push_back(q->val);
            if(!st.empty()) {
                TreeNode *q2 = st.top();
                while(q2->right && q2->right != q) {
                    st.push(q2->right); q2 = q2->right;
                    while(q2->left || q2->right) {
                        while(q2->left){ st.push(q2->left); q2 = q2->left;}
                        if(q2->right){ st.push(q2->right); q2 = q2->right;}
                    }   
                }
            }
        }
        return answer;
    }
};

Binary Tree Preorder Traversal

Given a binary tree, return the preorder traversal of its nodes' values.

For example: Given binary tree {1,#,2,3},

   1
    
     2
    /
   3

return [1,2,3].

Note: Recursive solution is trivial, could you do it iteratively?

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> preorderTraversal(TreeNode *root) {
        vector<int> vec;
        if(root == NULL) return vec;
        stack<TreeNode*> st;
        st.push(root);
        while(!st.empty()) {
            TreeNode *p = st.top(); st.pop();
            vec.push_back(p->val);
            if(p->right) st.push(p->right);
            if(p->left) st.push(p->left);
        }
        return vec;
    }
};