inorder,(long)Integer.MAX_VALUE+1);}intpre=0;intin=0;privateTreeNodebuildTreeHelper(int[]preorder,int[]inorder,longstop){//到达末尾返回 nullif(pre==preorder.length){returnnull;}//到达停止点返回 null//当前停止点已经用了,in 后移if(inorder[in]==stop){in++;returnnull;}introot_val=pre...
}privateTreeNode buildTree(int[] preorder,intpreLo,intpreHi,int[] inorder,intinLo,intinHi) {if(preLo > preHi || inLo >inHi)returnnull; TreeNode root=newTreeNode(preorder[preLo]);introotAtInorder = 0;for(inti = inLo; i <= inHi; i++) {if(inorder[i] == root.val) {//...
i=inorder.index(preorder[0]) root.left= self.buildTree(preorder[1:1+i], inorder[:i]) root.right= self.buildTree(preorder[i+1:], inorder[i+1:])returnroot 同样的方法做106. Construct Binary Tree from Inorder and Postorder Traversal,后序和中序,建二叉树,根节点是后序的最后一个元素。
public TreeNode buildTree(int[] preorder, int[] inorder) { if(preorder==null || preorder.length<=0 || inorder==null || inorder.length<=0 || preorder.length!=inorder.length) return null; else return getRootByRecursion(preorder,inorder,0,0,inorder.length-1); } /* * inStart和...
LeetCode: 105. Construct Binary Tree from Preorder and Inorder Traversal 题目描述 Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. For example, given ...
5. Construct Binary Tree from Preorder and Inorder Traversal 与上题类似,我们知道preorder第一个元素一定是根节点,按照这个特性可以找出根节点在inorder中的位置,从该位置分开,左右两边分别就是左右子树。 class Solution: def buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode: if...
Binary treeIn most of the Data Structure textbooks there is a conclusion: "Given the preoder and the inorder sequence of nodes, the binary tree structure can be determined exclusively". This paper gives a counter example to prove this conclusion is wrong, and analzes the reason which cause...
order B. Traversing the forest corresponding to the binary tree in root-last order C. Traversing the forest corresponding to the binary tree in breadth-first order D. None of the above 答案：A 分析：正确答案:A 解析:前序遍历一个二叉树等价于按从树的根部、右子树、右子树查找顺序查找树。
题目:105. Construct Binary Tree from Preorder and Inorder Traversal Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 翻译:给定树的前序和中序遍历,构造二叉树。 注意: 树中不存在重复项。
* TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */publicclassSolution{public TreeNodebuildTree(int[]preorder,int[]inorder){if(preorder==null||preorder.length==0)returnnull;if(inorder==null||inorder.length==0)returnnull;if(preorder.length!=inorder.length...