Binary Tree Maximum Path Sum
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Problem:
Given a binary tree, find the maximum path sum.
The path may start and end at any node in the tree.
For example:
Given the below binary tree,
1 / \ 2 3
Return 6
.
似乎没有比递归更优的解了。
pmax表示从根节点出发的最大路径和。
smax表示包含根节点的最大路径和。
pmax(root)=max(root.val,root.val+pmax(root.left),root.val+pmax(root.right))。
smax(root)=max(pmax(root),root.val+pmax(root.left)+pmax(root.right))。
Solution:
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
int MaxSum = Integer.MIN_VALUE;
public int maxPathSum(TreeNode root) {
if(root!=null)
subMaxSum(root);
return MaxSum;
}
private int subMaxSum(TreeNode root)
{
int lmax,rmax,smax,pmax;
if(root.left==null&&root.right==null)
{
smax = root.val;
pmax = root.val;
}
else if(root.left==null)
{
rmax = subMaxSum(root.right);
pmax = Math.max(root.val, root.val+rmax);
smax = pmax;
}
else if(root.right==null)
{
lmax = subMaxSum(root.left);
pmax = Math.max(root.val, root.val+lmax);
smax = pmax;
}
else
{
rmax = subMaxSum(root.right);
lmax = subMaxSum(root.left);
pmax = Math.max(root.val,Math.max(root.val+rmax,root.val+lmax));
smax = Math.max(pmax,root.val+lmax+rmax);
}
if(MaxSum<smax)
MaxSum = smax;
return pmax;
}
}
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
int MaxSum = Integer.MIN_VALUE;
public int maxPathSum(TreeNode root) {
if(root!=null)
subMaxSum(root);
return MaxSum;
}
private int subMaxSum(TreeNode root)
{
int lmax,rmax,smax,pmax;
if(root.left==null&&root.right==null)
{
smax = root.val;
pmax = root.val;
}
else if(root.left==null)
{
rmax = subMaxSum(root.right);
pmax = Math.max(root.val, root.val+rmax);
smax = pmax;
}
else if(root.right==null)
{
lmax = subMaxSum(root.left);
pmax = Math.max(root.val, root.val+lmax);
smax = pmax;
}
else
{
rmax = subMaxSum(root.right);
lmax = subMaxSum(root.left);
pmax = Math.max(root.val,Math.max(root.val+rmax,root.val+lmax));
smax = Math.max(pmax,root.val+lmax+rmax);
}
if(MaxSum<smax)
MaxSum = smax;
return pmax;
}
}
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