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# 662.Maximum-Width-of-Binary-Tree

## 题目描述

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.

## 代码

### Approach #1 DFS

Time Complexity & Space Complexity: O(N)
class Solution {
int ans;
Map<Integer, Integer> left;
public int widthOfBinaryTree(TreeNode root) {
ans = 0;
left = new HashMap();
dfs(root, 0, 0);
return ans;
}
public void dfs(TreeNode root, int depth, int pos) {
if (root == null) return;
left.putIfAbsent(depth, pos);
ans = Math.max(ans, pos - left.get(depth) + 1);
dfs(root.left, depth + 1, 2 * pos);
dfs(root.right, depth + 1, 2 * pos + 1);
}
}

### Approach #2 BFS

Time Complexity & Space Complexity: O(N)
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int widthOfBinaryTree(TreeNode root) {
int ans = 0;
int curDepth = 0;
int left = 0;
while (!queue.isEmpty()) {
AnnotateNode a = queue.poll();
if (a.node != null) {
queue.add(new AnnotateNode(a.node.left, a.depth + 1, a.pos * 2));
queue.add(new AnnotateNode(a.node.right, a.depth + 1, a.pos * 2 + 1));
if (curDepth != a.depth) {
curDepth = a.depth;
left = a.pos;
}
ans = Math.max(ans, a.pos - left + 1);
}
}
return ans;
}
}
class AnnotateNode {
TreeNode node;
int depth;
int pos;
AnnotateNode(TreeNode, n, int d, int p) {
node = n;
depth = d;
pos = p;
}
}

### Approach #3 Queue

class Solution {
public int widthOfBinaryTree(TreeNode root) {
if (root == null) return 0;
Map<TreeNode, Integer> m = new HashMap<TreeNode, Integer>();
q.offer(root);
m.put(root, 1);
int curW = 0;
int maxW = 0;
while (!q.isEmpty()) {
int size = q.size();
int start = 0;
int end = 0;
for (int i = 0; i < size; i++) {
TreeNode node = q.poll();
if (i == 0) start = m.get(node);
if (i == size - 1) end = m.get(node);
if (node.left != null) {
m.put(node.left, m.get(node) * 2);
q.offer(node.left);
}
if (node.right != null) {
m.put(node.right, m.get(node) * 2 + 1);
q.offer(node.right);
}
}
curW = end - start + 1;
maxW = Math.max(curW, maxW);
}
return maxW;
}
}