二叉樹的直徑（左右子樹的深度和）Diameter of Binary Tree

題目：

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.

解決：

①  題目要求求出二叉樹的直徑：二叉樹中從一個結點到另外一個節點最長的路徑，叫作二叉樹的直徑

咱們能夠理解爲求根節點的左右子樹的深度和，那麼咱們只要對每個結點求出其左右子樹深度之和，就能夠更新結果diameter了。爲了減小重複計算，咱們用hashmap創建每一個結點和其深度之間的映射，這樣某個結點的深度以前計算過了，就不用再次計算了。

/**
 * 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)));//不僅是頭節點
    }
    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 思路和上一個相同，可是要簡潔的多，沒有使用map，只使用一個遞歸便可。在求深度的遞歸函數中順便就把直徑算出來了。

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 進化版。

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 比較好理解的

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;     } }