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为了解决这个问题，我们需要找到一棵树中最远两点之间的距离，即树的直径。直径定义为路径上的节点值之和的最大值。

方法思路

为了找到树的直径，我们可以采用以下步骤：

解决代码

import java.util.ArrayList;import java.util.List;public class BinaryTreeDiameter {    public class TreeNode {        int val;        TreeNode left;        TreeNode right;        public TreeNode(int val) {            this.val = val;        }    }    public int diameterOfBinaryTree(TreeNode root) {        if (root == null) {            return 0;        }        int height = calculateTreeHeight(root);        int k = findCenterLevel(height);        List
   
     centerNodes = getCenterNodes(root, k);        return computeMaxPathSum(centerNodes);    }    private int calculateTreeHeight(TreeNode root) {        if (root == null) {            return 0;        }        int height = 0;        bfs(root, 0, height);        return height;    }    private void bfs(TreeNode node, int currentLevel, int& height) {        if (node == null) {            return;        }        if (currentLevel > height) {            height = currentLevel;        }        if (node.left != null) {            bfs(node.left, currentLevel + 1, height);        }        if (node.right != null) {            bfs(node.right, currentLevel + 1, height);        }    }    private int findCenterLevel(int h) {        int k = 0;        while (Math.pow(2, k) <= h) {            k++;        }        k--; // 最大的k满足2^k <= h        return k;    }    private List
    
      getCenterNodes(TreeNode root, int k) {        List
     
       centerNodes = new ArrayList<>();        bfs(root, 0, 0, centerNodes, k);        return centerNodes;    }    private void bfs(TreeNode node, int currentLevel, int targetLevel, List
      
        centerNodes, int k) {        if (node == null) {            return;        }        if (currentLevel == targetLevel) {            centerNodes.add(node);            return;        }        if (currentLevel < targetLevel) {            if (node.left != null) {                bfs(node.left, currentLevel + 1, targetLevel, centerNodes, k);            }            if (node.right != null) {                bfs(node.right, currentLevel + 1, targetLevel, centerNodes, k);            }        }    }    private int computeMaxPathSum(List
       
         centerNodes) {        int maxSum = 0;        for (TreeNode centerNode : centerNodes) {            int currentMax = computePathSum(centerNode);            if (currentMax > maxSum) {                maxSum = currentMax;            }        }        return maxSum;    }    private int computePathSum(TreeNode node) {        if (node == null) {            return 0;        }        int leftSum = computePathSum(node.left);        int rightSum = computePathSum(node.right);        int maxChildSum = Math.max(leftSum, rightSum);        return node.val + maxChildSum;    }}
       
      
     
    
   

代码解释

通过以上步骤，我们可以高效地找到树的直径，即路径上的节点值之和的最大值。

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