[Leetcode][第106题][JAVA][ 从中序与后序遍历序列构造二叉树][分治][递归]

【问题描述】[中等]

【解答思路】

public class Solution {public TreeNode buildTree(int[] inorder, int[] postorder) {int inLen = inorder.length;int postLen = postorder.length;// 特判if (inLen != postLen) {throw new RuntimeException("输入错误");}return buildTree(inorder, 0, inLen - 1, postorder, 0, postLen - 1);}/*** 使用中序遍历序列 inorder 的子区间 [inLeft, inRight]* 与后序遍历序列 postorder 的子区间 [postLeft, postRight] 构建二叉树** @param inorder   中序遍历序列* @param inLeft    中序遍历序列的左边界* @param inRight   中序遍历序列的右边界* @param postorder 后序遍历序列* @param postLeft  后序遍历序列的左边界* @param postRight 后序遍历序列的右边界* @return 二叉树的根结点*/private TreeNode buildTree(int[] inorder, int inLeft, int inRight,int[] postorder, int postLeft, int postRight) {if (inLeft > inRight || postLeft > postRight) {return null;}int pivot = postorder[postRight];int pivotIndex = inLeft;// 注意这里如果编写不当，有数组下标越界的风险while (inorder[pivotIndex] != pivot) {pivotIndex++;}TreeNode root = new TreeNode(pivot);root.left = buildTree(inorder, inLeft, pivotIndex - 1,postorder, postLeft, postRight - inRight + pivotIndex - 1);root.right = buildTree(inorder, pivotIndex + 1, inRight,postorder, postRight - inRight + pivotIndex, postRight - 1);return root;}
}作者：liweiwei1419
链接：https://leetcode-cn.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/solution/hou-xu-bian-li-python-dai-ma-java-dai-ma-by-liwe-2/

HashMap优化

class Solution {HashMap<Integer,Integer> memo = new HashMap<>();int[] post;public TreeNode buildTree(int[] inorder, int[] postorder) {for(int i = 0;i < inorder.length; i++) memo.put(inorder[i], i);post = postorder;TreeNode root = buildTree(0, inorder.length - 1, 0, post.length - 1);return root;}public TreeNode buildTree(int is, int ie, int ps, int pe) {if(ie < is || pe < ps) return null;int root = post[pe];int ri = memo.get(root);TreeNode node = new TreeNode(root);node.left = buildTree(is, ri - 1, ps, ps + ri - is - 1);node.right = buildTree(ri + 1, ie, ps + ri - is, pe - 1);return node;}
}作者：reals
链接：https://leetcode-cn.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/solution/tu-jie-gou-zao-er-cha-shu-wei-wan-dai-xu-by-user72/

copyOfRange 左必右开

class Solution {public TreeNode buildTree(int[] inorder, int[] postorder) {if(inorder==null || postorder==null) {return null;}return helper(inorder,postorder);}private TreeNode helper(int[] in, int[] post) {if(in.length==0) {return null;}//根据后序数组的最后一个元素，创建根节点TreeNode root = new TreeNode(post[post.length-1]);//在中序数组中查找值等于【后序数组最后一个元素】的下标for(int i=0;i<in.length;++i) {if(in[i]==post[post.length-1]) {//确定这个下标i后，将中序数组分成两部分，后序数组分成两部分int[] inLeft = Arrays.copyOfRange(in,0,i);int[] inRight = Arrays.copyOfRange(in,i+1,in.length);int[] postLeft = Arrays.copyOfRange(post,0,i);int[] postRight = Arrays.copyOfRange(post,i,post.length-1);//递归处理中序数组左边，后序数组左边root.left = helper(inLeft,postLeft);//递归处理中序数组右边，后序数组右边root.right = helper(inRight,postRight);break;}}return root;}
}作者：wang_ni_ma
链接：https://leetcode-cn.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/solution/liang-chong-shi-xian-dong-hua-yan-shi-106-cong-zho/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

【总结】

1.前中后序遍历变化的是[中]的位置，左到右的顺序不改变

前序遍历中左右
中序遍历左中右
后续遍历左右中

2.还原二叉树借助HashMap or copyOfRange

根据前序和后序遍历构造二叉树
[Leetcode][第889题][JAVA][根据前序和后序遍历构造二叉树][分治][递归]
前序+中序遍历可画出原二叉树
[Leedcode][JAVA][第105题][从前序与中序遍历序列构造二叉树][栈][递归][二叉树]
后续+中序遍历可画出原二叉树
[Leetcode][第106题][JAVA][ 从中序与后序遍历序列构造二叉树][分治][递归]

3. 多画图写写写遍历代码手撕变量大脑保持清醒

转载链接：https://leetcode-cn.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/solution/hou-xu-bian-li-python-dai-ma-java-dai-ma-by-liwe-2/
转载链接：https://leetcode-cn.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/solution/tu-jie-gou-zao-er-cha-shu-wei-wan-dai-xu-by-user72/