145.Binary Tree Postorder Traversal
Given the root of a binary tree, return the postorder traversal of its nodes’ values.
Example 1:
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| Input: root = [1,null,2,3]
Output: [3,2,1]
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Example 2:
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| Input: root = []
Output: []
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Example 3:
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| Input: root = [1]
Output: [1]
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递归法：
时间复杂度：O(n)
空间复杂度：O(n)
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| class Solution {
public:
void postorder(TreeNode* root, vector<int> &ans_v) {
if (root == nullptr)return;
else {
postorder(root->left,ans_v);
postorder(root->right,ans_v);
ans_v.push_back(root->val);
}
}
vector<int> postorderTraversal(TreeNode* root) {
vector<int> ans_v;
postorder(root,ans_v);
return ans_v;
}
};
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迭代法：
时间复杂度：O(n)
空间复杂度：O(n)
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| class Solution {
public:
vector<int> postorderTraversal(TreeNode *root) {
vector<int> res;
if (root == nullptr) {
return res;
}
stack<TreeNode *> stk;
TreeNode *prev = nullptr;
while (root != nullptr || !stk.empty()) {
while (root != nullptr) {
stk.emplace(root);
root = root->left;
}
root = stk.top();
stk.pop();
if (root->right == nullptr || root->right == prev) {
res.emplace_back(root->val);
prev = root;
root = nullptr;
} else {
stk.emplace(root);
root = root->right;
}
}
return res;
}
};
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Morris遍历：
时间复杂度：O(n)
空间复杂度：O(1)
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| class Solution {
public:
void addPath(vector<int> &vec, TreeNode *node) {
int count = 0;
while (node != nullptr) {
++count;
vec.emplace_back(node->val);
node = node->right;
}
reverse(vec.end() - count, vec.end());
}
vector<int> postorderTraversal(TreeNode *root) {
vector<int> res;
if (root == nullptr) {
return res;
}
TreeNode *p1 = root, *p2 = nullptr;
while (p1 != nullptr) {
p2 = p1->left;
if (p2 != nullptr) {
while (p2->right != nullptr && p2->right != p1) {
p2 = p2->right;
}
if (p2->right == nullptr) {
p2->right = p1;
p1 = p1->left;
continue;
} else {
p2->right = nullptr;
addPath(res, p1->left);
}
}
p1 = p1->right;
}
addPath(res, root);
return res;
}
};
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