LeetCode | 145.二叉树的后序遍历
给你一棵二叉树的根节点 root ,返回其节点值的 后序遍历 。
示例 1:
1
\
2
/
3
输入:root = [1,null,2,3]
输出:[3,2,1]
示例 2:
输入:root = []
输出:[]
示例 3:
输入:root = [1]
输出:[1]
提示:
- 树中节点的数目在范围 [0, 100] 内
- -100 <= Node.val <= 100
迭代:
vector<int> postorderTraversal(TreeNode* root) {
if (!root)
return {};
vector<int> res;
stack<TreeNode*> st;
TreeNode* pre = nullptr;
while (root || !st.empty()) {
while (root) {
st.push(root);
root = root->left;
}
root = st.top();
st.pop();
if (!root->right || root->right == pre) {
res.push_back(root->val);
pre = root;
root = nullptr;
} else {
st.push(root);
root = root->right;
}
}
return res;
}
递归:
void helper(TreeNode* root, vector<int> &res) {
if (!root)
return;
helper(root->left, res);
helper(root->right, res);
res.push_back(root->val);
}
vector<int> postorderTraversal(TreeNode* root) {
vector<int> res;
helper(root, res);
return res;
}
mirrors 算法:
设当前节点为 cur
没有左节点
直接遍历右节点
有左节点
找到左节点的最右的子节点
如果最右节点的右节点为空 (最右节点的右节点不一定为空,因为我们会令它指向 cur 节点)
令右节点指向 cur, 遍历 cur 的左节点
如果最右节点的右节点不为空,说明 cur 的左节点已经遍历完
断开最右节点和 cur 的连接,开始遍历 cur 的右节点
后序遍历比前序和中序要复杂一点,要遍历完左右节点才添加中间节点
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 *cur = root, *rightmost = nullptr;
while (cur != nullptr) {
if (cur->left != nullptr) {
rightmost = cur->left;
while (rightmost->right != nullptr && rightmost->right != cur) {
rightmost = rightmost->right;
}
if (rightmost->right == nullptr) { // 左节点还没有遍历
rightmost->right = cur;
cur = cur->left;
continue;
} else { // 左节点遍历完
rightmost->right = nullptr;
addPath(res, cur->left); // 左节点遍历完后添加到数组
}
}
cur = cur->right;
}
addPath(res, root);
return res;
}
标签:145,cur,res,LeetCode,right,二叉树,遍历,root,节点
From: https://www.cnblogs.com/AngleLin/p/17282799.html