Problem
We are given the root node of a maximum tree: a tree where every node has a value greater than any other value in its subtree.
Just as in the previous problem, the given tree was constructed from an list A (root = Construct(A)) recursively with the following Construct(A) routine:
If A is empty, return null.
Otherwise, let A[i] be the largest element of A. Create a root node with value A[i].
The left child of root will be Construct([A[0], A[1], …, A[i-1]])
The right child of root will be Construct([A[i+1], A[i+2], …, A[A.length - 1]])
Return root.
Note that we were not given A directly, only a root node root = Construct(A).
Suppose B is a copy of A with the value val appended to it. It is guaranteed that B has unique values.
Return Construct(B).
Example1
Input: root = [4,1,3,null,null,2], val = 5
Output: [5,4,null,1,3,null,null,2]
Explanation: A = [1,4,2,3], B = [1,4,2,3,5]
Example2
Input: root = [5,2,4,null,1], val = 3
Output: [5,2,4,null,1,null,3]
Explanation: A = [2,1,5,4], B = [2,1,5,4,3]
Example3
Input: root = [5,2,3,null,1], val = 4
Output: [5,2,4,null,1,3]
Explanation: A = [2,1,5,3], B = [2,1,5,3,4]
Solution
通过中序遍历树,可以恢复出A,进而得到B,然后以B为输入,构造结果。
还可以直接在树上进行操作。
根据新加入的值是大于根节点还是小于根节点,分两种情况进行处理。
/**
* 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* insertIntoMaxTree(TreeNode* root, int val) {
if(!root)
return new TreeNode(val);
TreeNode * prev = NULL;
insertion(root,val,prev);
if(prev)
return prev;
else
return root;
}
void insertion(TreeNode* root, int val,TreeNode *&prev)
{
if(!root)
{
TreeNode * node = new TreeNode(val);
if(prev)
prev->right = node;
return;
}
int rootVal = root->val;
if(val > rootVal)
{
TreeNode *newRoot = new TreeNode(val);
newRoot->left = root;
if(prev)
prev->right = newRoot;
else
prev = newRoot;
}
else
{
insertion(root->right,val,root);
}
}
};