951. Flip Equivalent Binary Trees
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Medium
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For a binary tree T, we can define a flip operation as follows: choose any node, and swap the left and right child subtrees.
A binary tree X is flip equivalent to a binary tree Y if and only if we can make X equal to Y after some number of flip operations.
Given the roots of two binary trees root1 and root2, return true if the two trees are flip equivalent or false otherwise.
Example 1:
Flipped Trees Diagram
Input: root1 = [1,2,3,4,5,6,null,null,null,7,8], root2 = [1,3,2,null,6,4,5,null,null,null,null,8,7]
Output: true
Explanation: We flipped at nodes with values 1, 3, and 5.
Example 2:
Input: root1 = [], root2 = []
Output: true
Example 3:
Input: root1 = [], root2 = [1]
Output: false
Constraints:
The number of nodes in each tree is in the range [0, 100].
Each tree will have unique node values in the range [0, 99].
Solved
Medium
Topics
Companies
For a binary tree T, we can define a flip operation as follows: choose any node, and swap the left and right child subtrees.
A binary tree X is flip equivalent to a binary tree Y if and only if we can make X equal to Y after some number of flip operations.
Given the roots of two binary trees root1 and root2, return true if the two trees are flip equivalent or false otherwise.
Example 1:
Flipped Trees Diagram
Input: root1 = [1,2,3,4,5,6,null,null,null,7,8], root2 = [1,3,2,null,6,4,5,null,null,null,null,8,7]
Output: true
Explanation: We flipped at nodes with values 1, 3, and 5.
Example 2:
Input: root1 = [], root2 = []
Output: true
Example 3:
Input: root1 = [], root2 = [1]
Output: false
Constraints:
The number of nodes in each tree is in the range [0, 100].
Each tree will have unique node values in the range [0, 99].
Class Solution {
Public Boolean Flipequiv(Treenode Root1, Treenode Root2) {
Return Dfs(Root1, Root2);
}
Private Boolean Dfs(Treenode Root1, Treenode Root2) {
If (Root1 == Root2 || (Root1 == Null && Root2 == Null)) {
Return True;
}
If (Root1 == Null Root2 == Null Root1.Val != Root2.Val) {
Return False;
}
Return (Dfs(Root1.Left, Root2.Left) && Dfs(Root1.Right, Root2.Right))
|| (Dfs(Root1.Left, Root2.Right) && Dfs(Root1.Right, Root2.Left));
}
}
Public Boolean Flipequiv(Treenode Root1, Treenode Root2) {
Return Dfs(Root1, Root2);
}
Private Boolean Dfs(Treenode Root1, Treenode Root2) {
If (Root1 == Root2 || (Root1 == Null && Root2 == Null)) {
Return True;
}
If (Root1 == Null
Return False;
}
Return (Dfs(Root1.Left, Root2.Left) && Dfs(Root1.Right, Root2.Right))
|| (Dfs(Root1.Left, Root2.Right) && Dfs(Root1.Right, Root2.Left));
}
}