题目

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 the null nodes 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.

计算二叉树每个节点的偏移值

1                      |1|                     d = 0, size = 2^0 = 1
/ \
3   2                    |3|2|                   d = 1, size = 2^1 = 2
/     \
5       9                  |5| | |9|               d = 2, size = 2^2 = 4
/         \
6           7                |6| | | | | | |7|       d = 3, size = 2^3 = 8
-------------------------------------------------
0 1 2 3 4 5 6 7

if offset(parent) = n, offset(left_child) = n * 2, offset(right_child) = n * 2 + 1

代码

class Solution {
public int widthOfBinaryTree(TreeNode root) {
Map<Integer, int[]> map = new HashMap<>();
parseTree(root, 0, 0, map);
int maxScope = 0;
for (Map.Entry<Integer, int[]> entry : map.entrySet()) {
int[] scope = entry.getValue();
maxScope = Math.max(maxScope, scope - scope + 1);
}
return maxScope;
}

private void parseTree(TreeNode node, int depth, int offset, Map<Integer, int[]> map) {
if (node != null) {
if (!map.containsKey(depth)) map.put(depth, new int[]{Integer.MAX_VALUE, Integer.MIN_VALUE});
int[] scope = map.get(depth);
scope = Math.min(scope, offset);
scope = Math.max(scope, offset);
parseTree(node.left, depth + 1, offset * 2, map);
parseTree(node.right, depth + 1, offset * 2 + 1, map);
}
}
}

结果 