Diameter of Binary Tree

题目：

Given a binary tree, you need to compute the length of the diameter of the tree. The diameter of a binary tree is the length of the longest path between any two nodes in a tree. This path may or may not pass through the root.

Example:
Given a binary tree

1
/ \
2   3
/ \
4   5

Return 3, which is the length of the path [4,2,1,3] or [5,2,1,3].

Note: The length of path between two nodes is represented by the number of edges between them.

解决：

① 题目要求求出二叉树的直径：二叉树中从一个结点到另一个节点最长的路径，叫做二叉树的直径

我们可以理解为求根节点的左右子树的深度和，那么我们只要对每一个结点求出其左右子树深度之和，就可以更新结果diameter了。为了减少重复计算，我们用hashmap建立每个结点和其深度之间的映射，这样某个结点的深度之前计算过了，就不用再次计算了。

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

public class Solution {//321 ms
public int diameterOfBinaryTree(TreeNode root) {
if(root == null) return 0;
int diameter = getDepth(root.left) + getDepth(root.right);
return Math.max(diameter,
Math.max(diameterOfBinaryTree(root.left),diameterOfBinaryTree(root.right)));//不只是头节点
}
public int getDepth(TreeNode node){
Map<TreeNode,Integer> map = new HashMap<>();
if(node == null) return 0;
if(map.containsKey(node) && map.get(node) > 0) return map.get(node);
int depth = Math.max(getDepth(node.left),getDepth(node.right)) + 1;
map.put(node,depth);
return map.get(node);
}
}

② 思路和上一个相同，但是要简洁的多，没有使用map，只使用一个递归即可。在求深度的递归函数中顺便就把直径算出来了。

public class Solution {//10ms
int diameter = 0;
public int diameterOfBinaryTree(TreeNode root) {
getDepth(root);
return diameter;
}
private int getDepth(TreeNode root) {
if(root == null) return 0;
int left = getDepth(root.left);
int right = getDepth(root.right);
diameter = Math.max(diameter, left + right); //不是left + right + 1。。是path。。不是number of node
return Math.max(left, right) + 1;
}
}

③ 进化版。

public class Solution {//9ms
int depth = 0;
public int diameterOfBinaryTree(TreeNode root) {
getDepth(root);
return depth;
}
public int getDepth(TreeNode root) {
if (root == null) return 0;
int left = getDepth(root.left);
int right = getDepth(root.right);
if (depth < left + right) {
depth = left + right;
}
return left > right ? left + 1 : right + 1;
}
}

④ 比较好理解的

class Solution {//23ms
public int diameterOfBinaryTree(TreeNode root) {
if(root == null) return 0;
int diameter = getDepth(root.left) + getDepth(root.right);
return Math.max(diameter,Math.max(diameterOfBinaryTree(root.left),diameterOfBinaryTree(root.right)));
}
public int getDepth(TreeNode root){
if(root == null) return 0;
return Math.max(getDepth(root.left),getDepth(root.right)) + 1;
}
}