河南建设信息网一体化企业seo推广的绝密诀窍曝光

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前言

二叉树解题的思维模式分两类： 1、是否可以通过遍历一遍二叉树得到答案？如果可以，用一个 traverse 函数配合外部变量来实现，这叫「遍历」的思维模式。 2、是否可以定义一个递归函数，通过子问题（子树）的答案推导出原问题的答案？如果可以，写出这个递归函数的定义，并充分利用这个函数的返回值，这叫「分解问题」的思维模式。 无论使用哪种思维模式，你都需要思考： 如果单独抽出一个二叉树节点，它需要做什么事情？需要在什么时候（前/中/后序位置）做？其他的节点不用你操心，递归函数会帮你在所有节点上执行相同的操作。

一、力扣654. 最大二叉树

/*** 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 constructMaximumBinaryTree(int[] nums) {return fun(nums, 0, nums.length-1);}public TreeNode fun(int[] nums, int low, int high){if(low > high){return null;}int index = low;for(int i = low+1; i <= high; i ++){if(nums[i] > nums[index]){index = i;}}TreeNode cur = new TreeNode(nums[index]);TreeNode l = fun(nums,low,index-1);TreeNode r = fun(nums,index+1,high);cur.left = l;cur.right = r;return cur;}
}

二、力扣105. 从前序与中序遍历序列构造二叉树

/*** 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 buildTree(int[] preorder, int[] inorder) {return fun(preorder, inorder, 0, preorder.length-1, 0, inorder.length-1);}public TreeNode fun(int[] preorder, int[] inorder, int preLeft,int preRight,int inPre,int inRight){if(preLeft > preRight || inPre > inRight){return null;}TreeNode root = new TreeNode(preorder[preLeft]);int len = 0;for(int i = inPre; i <= inRight; i ++){if(inorder[i] == preorder[preLeft]){len = i - inPre;break;}}root.left = fun(preorder, inorder, preLeft+1, preLeft+len, inPre, inPre+len-1);root.right = fun(preorder, inorder, preLeft+len+1,preRight, inPre+len+1,inRight);return root;}
}

三、力扣106. 从中序与后序遍历序列构造二叉树

/*** 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 {Map<Integer,Integer> invail;public TreeNode buildTree(int[] inorder, int[] postorder) {invail = new HashMap<>();for(int i = 0; i < inorder.length; i ++){invail.put(inorder[i],i);}return fun(inorder, postorder, 0, postorder.length-1,0,inorder.length-1);}public TreeNode fun(int[] inorder, int[] postorder, int postLeft, int postRight, int inLeft, int inRight){if(postLeft > postRight || inLeft > inRight){return null;}TreeNode root = new TreeNode(postorder[postRight]);int index = invail.get(postorder[postRight]);int len = index - inLeft;root.left = fun(inorder, postorder, postLeft, postLeft+len-1,inLeft, index-1);root.right = fun(inorder, postorder, postLeft+len,postRight-1,index+1,inRight);return root;}
}

四、力扣889. 根据前序和后序遍历构造二叉树

/*** 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 {Map<Integer,Integer> invail;public TreeNode constructFromPrePost(int[] preorder, int[] postorder) {invail = new HashMap<>();for(int i = 0; i < postorder.length; i ++){invail.put(postorder[i],i);}return fun(preorder,postorder,0,preorder.length-1,0,postorder.length-1);}public TreeNode fun(int[] preorder, int[] postorder, int preLeft, int preRight,int postLeft,int postRight){if(preLeft > preRight || postLeft > postRight){return null;}TreeNode cur = new TreeNode(preorder[preLeft]);if(preLeft == preRight)return cur;int index = invail.get(preorder[preLeft+1]);int len = index - postLeft+1;cur.left = fun(preorder, postorder, preLeft+1, preLeft+len, postLeft, index);cur.right = fun(preorder, postorder, preLeft+len+1,preRight,index+1,postRight-1);return cur;}
}