【LeetCode】94. Binary Tree Inorder Traversal(C++)
阿新 • • 發佈:2018-12-19
地址:https://leetcode.com/problems/binary-tree-inorder-traversal/
題目:
Given a binary tree, return the inorder traversal of its nodes’ values.
Example:
理解:
二叉樹的中序遍歷
遞迴實現:
class Solution {
public:
vector<int> inorderTraversal(TreeNode* root) {
vector<int> res;
InOrder(root, res);
return res;
}
void InOrder(TreeNode* root, vector<int>& res) {
if (!root) return;
InOrder(root->left, res);
res.push_back(root->val);
InOrder(root->right, res);
}
};
非遞迴實現:
class Solution {
public:
vector<int> inorderTraversal(TreeNode* root) {
vector< int> res;
stack<TreeNode*> stk;
TreeNode* p = root;
while (p || !stk.empty()) {
if (p) {
stk.push(p);
p = p->left;
}
else {
p= stk.top();
stk.pop();
res.push_back(p->val);
p = p->right;
}
}
return res;
}
};
基於線索樹的實現
前面兩種空間複雜度都是
基於線索數,可以把空間複雜度變為
class Solution {
public:
vector<int> inorderTraversal(TreeNode* root) {
vector<int> res;
TreeNode* curr = root;
while (curr) {
if (curr->left) {
TreeNode* pre = curr->left;
while (pre->right && (pre->right != curr)) {
pre = pre->right;
}
//如果是第一次到達,修改線索樹
if (!pre->right) {
pre->right = curr;
curr = curr->left;
}
//第二次到達,改回去
else {
pre->right = nullptr;
res.push_back(curr->val);
curr = curr->right;
}
}
else {
res.push_back(curr->val);
curr = curr->right;
}
}
return res;
}
};