Given a binary tree, write a function to get the maximum width of the given tree. The width of a tree is the maximum width among all levels. The binary tree has the same structure as a full binary tree, but some nodes are null.
The width of one level is defined as the length between the end-nodes (the leftmost and right most non-null nodes in the level, where thenullnodes between the end-nodes are also counted into the length calculation.
Example 1:
Input:
1
/ \
3 2
/ \ \
5 3 9
Output:
4
Explanation:
The maximum width existing in the third level with the length 4 (5,3,null,9).
Example 2:
Input:
1
/
3
/ \
5 3
Output:
2
Explanation:
The maximum width existing in the third level with the length 2 (5,3).
Example 3:
Input:
1
/ \
3 2
/
5
Output:
2
Explanation:
The maximum width existing in the second level with the length 2 (3,2).
Example 4:
Input:
1
/ \
3 2
/ \
5 9
/ \
6 7
Output:
8
Explanation:
The maximum width existing in the fourth level with the length 8 (6,null,null,null,null,null,null,7).
Note:Answer will in the range of 32-bit signed integer.
2. Implementation
(1) BFS
class Solution {
class NodeInfo {
TreeNode node;
int pos;
public NodeInfo(TreeNode node, int pos) {
this.node = node;
this.pos = pos;
}
}
public int widthOfBinaryTree(TreeNode root) {
if (root == null) {
return 0;
}
Queue<NodeInfo> queue = new LinkedList<>();
queue.add(new NodeInfo(root, 1));
int start = 0, end = 0, size = 0, maxWidth = 0;
while (!queue.isEmpty()) {
size = queue.size();
start = queue.peek().pos;
for (int i = 0; i < size; i++) {
NodeInfo curNodeInfo = queue.remove();
end = curNodeInfo.pos;
if (curNodeInfo.node.left != null) {
queue.add(new NodeInfo(curNodeInfo.node.left, 2 * end));
}
if (curNodeInfo.node.right != null) {
queue.add(new NodeInfo(curNodeInfo.node.right, 2 * end + 1));
}
}
maxWidth = Math.max(maxWidth, end - start + 1);
}
return maxWidth;
}
}