LeetCode124. Binary Tree Maximum Path Sum
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LeetCode124. Binary Tree Maximum Path Sum
找出二叉树任意一点到另一点的路径,使得和最大.
解题思路:后序遍历,先计算左右子树的值l和r,若l<0或r<0,则不用加上
1 / \ 2 3
最大为6
4 / \ 2 6 / \ / \ 1 3 5 7
最大从3到7:4 + 5 + 13 = 22
public class Solution { int maxsum = Integer.MIN_VALUE; public int maxPathSum(TreeNode root) { if(root == null) return 0; maxSum(root); return maxsum; } public int maxSum(TreeNode root) { if(root == null) return 0; /*求以root为根的当前子树的最大路径和*/ int cur = root.val; int lmax = maxSum(root.left); int rmax = maxSum(root.right); if(lmax > 0) cur += lmax; if(rmax > 0) cur += rmax; if(cur > maxsum) maxsum = cur; /*返回以当前root为根的子树的最大路径和*/ return Math.max(root.val,Math.max(root.val + lmax, root.val + rmax)); }}
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