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【LeetCode】Binary Tree Inorder Traversal (2 solutions)
Binary Tree Inorder Traversal
Given a binary tree, return the inorder traversal of its nodes‘ values.
For example:
Given binary tree {1,#,2,3},
1 2 / 3
return [1,3,2].
Note: Recursive solution is trivial, could you do it iteratively?
解法一：递归
/** * 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> inorderTraversal(TreeNode *root) { vector<int> res; if(root == NULL) return res; inorder(root, res); return res; } void inorder(TreeNode* root, vector<int>& res) { if(root->left) inorder(root->left, res); res.push_back(root->val); if(root->right) inorder(root->right, res); }};
解法二：非递归
使用map记录是否访问过，使用栈记录访问路径，遵循左根右原则。
/** * 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> inorderTraversal(TreeNode *root) { vector<int> res; if(root == NULL) return res; stack<TreeNode*> s; map<TreeNode*, bool> m; s.push(root); m[root] = true; while(root->left) { s.push(root->left); m[root->left] = true; root = root->left; } while(!s.empty()) { TreeNode* top = s.top(); TreeNode* cur = top; //left bool tag = false; while(cur->left && m.find(cur->left)==m.end()) { tag = true; s.push(cur->left); m[cur->left] = true; cur = cur->left; } if(tag) continue; //root s.pop(); res.push_back(top->val); //right if(top->right) { s.push(top->right); m[top->right] = true; } } return res; }};
【LeetCode】Binary Tree Inorder Traversal (2 solutions)
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