LeetCode-----Construct Binary Tree from Preorder and Inorder Traversal
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2022-04-27 11:49:04
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Construct Binary Tree from Preorder and Inorder Traversal
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
由于先序的顺序的第一个肯定是根,所以原二叉树的根节点可以知道,题目中给了一个很关键的条件就是树中没有相同元素,有了这个条件我们就可以在中序遍历中也定位出根节点的位置,并以根节点的位置将中序遍历拆分为左右两个部分,分别对其递归调用原函数。
代码如下:
package com.zhumq.lianxi;
public class PreorderAndInorder {
//定义TreeNode
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) { val = x; }
}
public TreeNode buildTree(int[] preorder, int[] inorder) {
return buildTree(preorder, 0, preorder.length-1,inorder, 0, inorder.length-1);
}
public TreeNode buildTree(int[] preorder, int pleft, int pright,int[] inorder, int ileft, int iright) {
if (ileft > iright || pleft > pright) return null;
//找到中序对应的根节点的索引,用i记录
int i = 0;
for (i = ileft; i < inorder.length; ++i) {
if (inorder[i] == preorder[pleft]) break;
}
//初始根节点为前序的第一个元素
TreeNode cur = new TreeNode(preorder[pleft]);
//分成左右子树两个问题,两子树的根节点分别为cur.left和cur.right
cur.left = buildTree(preorder, pleft + 1, pleft + i - ileft, inorder, ileft, i - 1);
cur.right = buildTree(preorder, pleft + i - ileft + 1, pright, inorder, i + 1, iright);
return cur;
}
}
下面来看一个例子, 某一二叉树的先序和中序遍历分别为:
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