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前言

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最常见的，二叉树的列构造问题一般都会用到分解问题的思维模式。

一、力扣100. 相同的树

/*** 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 boolean isSameTree(TreeNode p, TreeNode q) {if(p == null && q == null){return true;}if(p == null || q == null){return false;}if(p.val != q.val){return false;}return isSameTree(p.left,q.left) && isSameTree(p.right, q.right);}
}

二、力扣1367. 二叉树中的链表

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode() {}*     ListNode(int val) { this.val = val; }*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
/*** 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 boolean isSubPath(ListNode head, TreeNode root) {if(head == null){return true;}if(root == null){return false;}if(head.val == root.val){if(fun(root,head)){return true;}}return isSubPath(head,root.left) || isSubPath(head,root.right);}public boolean fun(TreeNode root, ListNode head){if(head == null){return true;}if(root == null){return false;}if(root.val != head.val){return false;}return fun(root.left, head.next) || fun(root.right, head.next);}
}

三、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 preLow,int preHigh, int inLow,int inHigh){if(preLow > preHigh){return null;}TreeNode cur = new TreeNode(preorder[preLow]);int index = 0, len = 0;for(int i = inLow; i <= inHigh; i ++){if(inorder[i] == preorder[preLow]){index = i;len = index - inLow;break;}}cur.left = fun(preorder,inorder,preLow+1,preLow+len,inLow,index-1);cur.right = fun(preorder,inorder,preLow+len+1,preHigh,index+1,inHigh);return cur;}
}

四、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; i <= high; i ++){index = nums[i] > nums[index] ? i:index;}TreeNode cur = new TreeNode(nums[index]);cur.left = fun(nums,low, index-1);cur.right = fun(nums,index+1,high);return cur;}
}