教做游戏的网站营业推广是什么

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前言

二叉树的递归分为「遍历」和「分解问题」两种思维模式，这道题需要用到「分解问题」的思维，而且涉及处理子树，需要用后序遍历

一、力扣865. 具有所有最深节点的最小子树

/*** 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 {public TreeNode subtreeWithAllDeepest(TreeNode root) {Result res = fun(root);return res.node;}public Result fun(TreeNode root){if(root == null){return new Result(null,0);}Result left = fun(root.left);Result right = fun(root.right);if(left.depth == right.depth){return new Result(root,left.depth+1);}Result res = left.depth > right.depth ? left : right;res.depth = res.depth + 1;return res;}
}
class Result{public TreeNode node;public int depth;public Result(TreeNode node, int depth){this.node = node;this.depth = depth;}
}

二、力扣1123. 最深叶节点的最近公共祖先

/*** 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 {public TreeNode lcaDeepestLeaves(TreeNode root) {Result res = fun(root);return res.node;}public Result fun(TreeNode root){if(root == null){return new Result(null,0);}Result left = fun(root.left);Result right = fun(root.right);if(left.depth == right.depth){return new Result(root,left.depth+1);}Result res = left.depth > right.depth ? left : right;res.depth = res.depth + 1;return res;}
}
class Result{public TreeNode node;public int depth;public Result(TreeNode node, int depth){this.node = node;this.depth = depth;}
}

三、力扣1026. 节点与其祖先之间的最大差值

/*** 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 res = 0;public int maxAncestorDiff(TreeNode root) {fun(root);return res;}public int[] fun(TreeNode root){if(root == null){return new int[]{Integer.MAX_VALUE,Integer.MIN_VALUE};}int[] leftMinMax = fun(root.left);int[] rightMinMax = fun(root.right);int curMin = Math.min(Math.min(leftMinMax[0],rightMinMax[0]),root.val);int curMax = Math.max(Math.max(leftMinMax[1],rightMinMax[1]),root.val);res = Math.max(res,Math.max(curMax - root.val, root.val - curMin));return new int[]{curMin,curMax};}
}

四、力扣1120. 子树的最大平均值

/*** 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 {double res = 0;public double maximumAverageSubtree(TreeNode root) {fun(root);return res;}public double[] fun(TreeNode root){if(root == null){return new double[]{0,0};}double[] left = fun(root.left);double[] right = fun(root.right);double curCount = left[0] + right[0] + 1;double curSum = left[1] + right[1] + root.val;res = Math.max(res,curSum/curCount);if(curCount == 1){return new double[]{curCount,root.val};}return new double[]{curCount,curSum};}
}