LeetCode.124. 二叉树中的最大路径和
难度:hard
递归:
/**
* 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 ans = Integer.MIN_VALUE;
public int maxPathSum(TreeNode root) {
maxGet(root);
return ans;
}
public int maxGet(TreeNode node) {
if (node == null) {
return 0;
}
int leftMax = Math.max(0, maxGet(node.left));
int rightMax = Math.max(0, maxGet(node.right));
ans = Math.max(ans, leftMax + rightMax + node.val);
return Math.max(leftMax, rightMax) + node.val;
}
}
复杂度分析:
- 时间复杂度:O(N),遍历N个节点
- 空间复杂度:O(N)递归调用的层数,最坏情况:二叉树的高度等于节点个数