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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; }};
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