LeetCode OJ 543 Diameter of Binary Tree [Easy]
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题目描述:
Given a binary tree,you need to compute the length of the diameter of the tree. The diameter of abinary 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 isthe length of the path [4,2,1,3] or [5,2,1,3].
题目理解:
给一个二叉树,计算二叉树的最长的路径
题目分析:
1. 对于每一个节点,通过该节点的最长路径 = 该节点左子数最大深度 + 该节点右子数最大深度;
2. 计算通过每一节点的最长路径,取最大值;
3. 使用递归,极限情况是子树为null,此时返回0,即子树的深度为0
4. 使用递归,非极限情况,返回的是左/右子树的深度,左/右子树的深度 = max(子树的左子树深度left,子树的右子树深度right)+1, 这里的1表示加上该子树的根节点。
5. 对下图的递归的解释:
1
/ \
2 3
/ \
4 5
\
6
4/6/3的左右子树深度为0;
5的左子树深度为0,右子树(6)深度为0+1=1;
2的左子树(4)深度为0+1=1,右子树(5)深度为 = max((5)的左子树深度0,(5)的右子树深度1)+1=2;
1右子树(3)的深度为0+1=1,的左子树(2)深度为 = max((2)的左子树深度1,(2)的右子树深度2)+1=3;
题目解答:
public class Solution { int max = 0; public int diameterOfBinaryTree(TreeNoderoot) { maxDepth(root); return max; } private int maxDepth(TreeNoderoot) { if (root== null) return 0; intleft = maxDepth(root.left); int right= maxDepth(root.right); max =Math.max(max, left + right); return Math.max(left,right) + 1; }}
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