662. Maximum Width of Binary Tree

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:

\
        3     2
       / \     \ 

Output: 4
Explanation: The maximum width existing in the third level with the length 4 (5,3,null,9).
Example 2:
Input:

\ 

Output: 2
Explanation: The maximum width existing in the third level with the length 2 (5,3).
Example 3:
Input:

\

Output: 2
Explanation: The maximum width existing in the second level with the length 2 (3,2).
Example 4:
Input:

\
        3   2
       /     \  
      5       9 
     /         \

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.

思路：
Depth-First Search
Intuition and Algorithm

Traverse each node in depth-first order, keeping track of that node’s position. For each depth, the position of the first node reached of that depth will be kept in left[depth].

Then, for each node, a candidate width is pos - left[depth] + 1. We take the maximum of the candidate answers.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
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.computeIfAbsent(depth, x-> 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);
    }
}