【leetcode hot 100 543】二叉树的直径

错误解法一：（递归）直径=左边的高度+右边的高度

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int diameterOfBinaryTree(TreeNode root) {
        // 左边的高度+右边的高度
        int heightLeft = height(root.left);
        int heightRight = height(root.right);
        return heightLeft+heightRight;
    }
    
    public int height(TreeNode root){
        if(root==null){
            return 0;
        }
        return Math.max(height(root.left), height(root.right))+1;
    }
}

错误原因：

解法一：（深度优先搜索）任意一条路径均可以被看作由某个节点为起点，从其左儿子和右儿子向下遍历的路径拼接得到。要循环所有节点，计算路径取最大值。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    int res;
    public int diameterOfBinaryTree(TreeNode root) {
        res = 0;
        height(root);
        return res;
    }
    
    public int height(TreeNode root){
        if(root==null){
            return 0;
        }
        // 左边的高度+右边的高度
        int left = height(root.left);
        int right = height(root.right);
        res = Math.max(res, left+right);
        return Math.max(left, right)+1;
    }
}

注意：

• +1要放在Math.max(left, right)+1上，而不是int left = height(root.left)+1和int right = height(root.right)+1上，会影响res = Math.max(res, left+right)的计算。+1是返回上一节点的高度，而不是这一节点的高度。