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。