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class Solution {
public:
vector<int> preorderTraversal(TreeNode* root) {
if (!root) return {};
vector<int> res;
stack<TreeNode*> s{{root}};
while (!s.empty()) {
TreeNode *t = s.top(); s.pop();
res.push_back(t->val);
if (t->right) s.push(t->right);
if (t->left) s.push(t->left);
}
return res;
}
};
下面这种写法使用了一个辅助结点p,这种写法其实可以看作是一个模版,对应的还有中序和后序的模版写法,形式很统一,方便于记忆。辅助结点p初始化为根结点,while 循环的条件是栈不为空或者辅助结点p不为空,在循环中首先判断如果辅助结点p存在,那么先将p加入栈中,然后将p的结点值加入结果 res 中,此时p指向其左子结点。否则如果p不存在的话,表明没有左子结点,取出栈顶结点,将p指向栈顶结点的右子结点,参见代码如下:
解法二:
class Solution {
public:
vector<int> preorderTraversal(TreeNode* root) {
vector<int> res;
stack<TreeNode*> st;
TreeNode *p = root;
while (!st.empty() || p) {
if (p) {
st.push(p);
res.push_back(p->val);
p = p->left;
} else {
p = st.top(); st.pop();
p = p->right;
}
}
return res;
}
};
Given a binary tree, return the preorder traversal of its nodes' values.
Example:
Follow up: Recursive solution is trivial, could you do it iteratively?
一般我们提到 树的遍历,最常见的有先序遍历,中序遍历,后序遍历和层序遍历,它们用递归实现起来都非常的简单。而题目的要求是不能使用递归求解,于是只能考虑到用非递归的方法,这就要用到stack来辅助运算。由于先序遍历的顺序是"根-左-右", 算法为:
1. 把根节点 push 到栈中
2. 循环检测栈是否为空,若不空,则取出栈顶元素,保存其值,然后看其右子节点是否存在,若存在则 push 到栈中。再看其左子节点,若存在,则 push 到栈中。
参见代码如下:
解法一:
下面这种写法使用了一个辅助结点p,这种写法其实可以看作是一个模版,对应的还有中序和后序的模版写法,形式很统一,方便于记忆。辅助结点p初始化为根结点,while 循环的条件是栈不为空或者辅助结点p不为空,在循环中首先判断如果辅助结点p存在,那么先将p加入栈中,然后将p的结点值加入结果 res 中,此时p指向其左子结点。否则如果p不存在的话,表明没有左子结点,取出栈顶结点,将p指向栈顶结点的右子结点,参见代码如下:
解法二:
Github 同步地址:
#144
类似题目:
Binary Tree Inorder Traversal
Binary Tree Postorder Traversal
Binary Tree Level Order Traversal
Verify Preorder Sequence in Binary Search Tree
Verify Preorder Serialization of a Binary Tree
参考资料:
https://leetcode.com/problems/binary-tree-preorder-traversal/
https://leetcode.com/problems/binary-tree-preorder-traversal/discuss/45468/3-Different-Solutions
https://leetcode.com/problems/binary-tree-preorder-traversal/discuss/45493/Accepted-code.-Explaination-with-Algo.
https://leetcode.com/problems/binary-tree-preorder-traversal/discuss/45266/Accepted-iterative-solution-in-Java-using-stack.
LeetCode All in One 题目讲解汇总(持续更新中...)
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