654. Maximum Binary Tree
2017-09-13 23:13
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原题如下:
Given an integer array with no duplicates. A maximum tree building on this array is defined as follow:
The root is the maximum number in the array.
The left subtree is the maximum tree constructed from left part subarray divided by the maximum number.
The right subtree is the maximum tree constructed from right part subarray divided by the maximum number.
Construct the maximum tree by the given array and output the root node of this tree.
Example 1:
Note:
The size of the given array will be in the range [1,1000].
题目的要求是根据给出的数组构建一颗二叉树,构建规则如下:
1.数组中的最大值为根节点
2.以根节点为准,左子树的根节点为数组左半部分的最大值
3.以根节点为准,右子树的根节点为数组右半部分的最大值
重复以上步骤,构建二叉树,最终返回二叉树的根节点。
我一开始的思路是直接利用递归解决,分别返回左右子树的根节点。但是这样做的话,感觉会比较麻烦,而且复杂度可能会比较高。看到讨论区有时间复杂度为O(n)的方法,根据题目所给的例子,尝试着去解决问题。
初始化一个vector<TreeNode*> root模拟栈的行为,来存放遍历数组产生的当前节点。遍历数组,生成当前节点add。
①当栈root为空时,节点add直接入栈。进入下一循环:
TreeNode* add = new TreeNode(nums[i]);
if (root.empty()) {
root.push_back(add);
}
②当栈不为空时,比较当前节点add与栈顶节点root.back()的val大小:
如果root.back的val小于当前节点add,说明add的left可能指向root.back;将root.back弹出栈,比较下一个元素,直到栈为空或者栈顶元素的val大于add,则将root.back的右节点指向add;最后将当前节点add入栈
while (!root.empty() && root.back()->val < nums[i]) {
add->left = root.back();
root.pop_back();
}
if (!root.empty()) root.back()->right = add;
root.push_back(add);
所以,经过上述步骤,最终树根节点会在栈低,用root.front()返回。
完整代码如下:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* constructMaximumBinaryTree(vector<int>& nums) {
vector<TreeNode*> root;
for (int i = 0; i < nums.size(); i++) {
TreeNode* add = new TreeNode(nums[i]);
if (root.empty()) {
root.push_back(add);
}
else {
while (!root.empty() && root.back()->val < nums[i]) {
add->left = root.back();
root.pop_back();
}
if (!root.empty()) root.back()->right = add;
root.push_back(add);
}
}
return root.front();
}
};
Given an integer array with no duplicates. A maximum tree building on this array is defined as follow:
The root is the maximum number in the array.
The left subtree is the maximum tree constructed from left part subarray divided by the maximum number.
The right subtree is the maximum tree constructed from right part subarray divided by the maximum number.
Construct the maximum tree by the given array and output the root node of this tree.
Example 1:
Input: [3,2,1,6,0,5] Output: return the tree root node representing the following tree: 6 / \ 3 5 \ / 2 0 \ 1
Note:
The size of the given array will be in the range [1,1000].
题目的要求是根据给出的数组构建一颗二叉树,构建规则如下:
1.数组中的最大值为根节点
2.以根节点为准,左子树的根节点为数组左半部分的最大值
3.以根节点为准,右子树的根节点为数组右半部分的最大值
重复以上步骤,构建二叉树,最终返回二叉树的根节点。
我一开始的思路是直接利用递归解决,分别返回左右子树的根节点。但是这样做的话,感觉会比较麻烦,而且复杂度可能会比较高。看到讨论区有时间复杂度为O(n)的方法,根据题目所给的例子,尝试着去解决问题。
初始化一个vector<TreeNode*> root模拟栈的行为,来存放遍历数组产生的当前节点。遍历数组,生成当前节点add。
①当栈root为空时,节点add直接入栈。进入下一循环:
TreeNode* add = new TreeNode(nums[i]);
if (root.empty()) {
root.push_back(add);
}
②当栈不为空时,比较当前节点add与栈顶节点root.back()的val大小:
如果root.back的val小于当前节点add,说明add的left可能指向root.back;将root.back弹出栈,比较下一个元素,直到栈为空或者栈顶元素的val大于add,则将root.back的右节点指向add;最后将当前节点add入栈
while (!root.empty() && root.back()->val < nums[i]) {
add->left = root.back();
root.pop_back();
}
if (!root.empty()) root.back()->right = add;
root.push_back(add);
所以,经过上述步骤,最终树根节点会在栈低,用root.front()返回。
完整代码如下:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* constructMaximumBinaryTree(vector<int>& nums) {
vector<TreeNode*> root;
for (int i = 0; i < nums.size(); i++) {
TreeNode* add = new TreeNode(nums[i]);
if (root.empty()) {
root.push_back(add);
}
else {
while (!root.empty() && root.back()->val < nums[i]) {
add->left = root.back();
root.pop_back();
}
if (!root.empty()) root.back()->right = add;
root.push_back(add);
}
}
return root.front();
}
};
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