From a2a2cf60272094c3c9d962723132502d93db097a Mon Sep 17 00:00:00 2001 From: lizhendong128 Date: Fri, 3 Jun 2022 21:42:36 +0800 Subject: [PATCH] =?UTF-8?q?=E4=BF=AE=E6=94=B90106=E4=BB=8E=E4=B8=AD?= =?UTF-8?q?=E5=BA=8F=E4=B8=8E=E5=90=8E=E5=BA=8F=E9=81=8D=E5=8E=86=E5=BA=8F?= =?UTF-8?q?=E5=88=97=E6=9E=84=E9=80=A0=E4=BA=8C=E5=8F=89=E6=A0=91=20Java?= =?UTF-8?q?=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 将0106和0105的Java版本进行了修改，采用了map来存储位置信息，加快定位；并且代码更容易看懂 --- ...序与后序遍历序列构造二叉树.md | 86 +++++++++---------- 1 file changed, 39 insertions(+), 47 deletions(-) diff --git a/problems/0106.从中序与后序遍历序列构造二叉树.md b/problems/0106.从中序与后序遍历序列构造二叉树.md index 188ad3cb..878c3572 100644 --- a/problems/0106.从中序与后序遍历序列构造二叉树.md +++ b/problems/0106.从中序与后序遍历序列构造二叉树.md @@ -584,35 +584,29 @@ tree2 的前序遍历是[1 2 3]， 后序遍历是[3 2 1]。 ```java class Solution { + Map map; // 方便根据数值查找位置 public TreeNode buildTree(int[] inorder, int[] postorder) { - return buildTree1(inorder, 0, inorder.length, postorder, 0, postorder.length); + map = new HashMap<>(); + for (int i = 0; i < inorder.length; i++) { // 用map保存中序序列的数值对应位置 + map.put(inorder[i], i); + } + + return findNode(inorder, 0, inorder.length, postorder,0, postorder.length); // 前闭后开 } - public TreeNode buildTree1(int[] inorder, int inLeft, int inRight, - int[] postorder, int postLeft, int postRight) { - // 没有元素了 - if (inRight - inLeft < 1) { + + public TreeNode findNode(int[] inorder, int inBegin, int inEnd, int[] postorder, int postBegin, int postEnd) { + // 参数里的范围都是前闭后开 + if (inBegin >= inEnd || postBegin >= postEnd) { // 不满足左闭右开，说明没有元素，返回空树 return null; } - // 只有一个元素了 - if (inRight - inLeft == 1) { - return new TreeNode(inorder[inLeft]); - } - // 后序数组postorder里最后一个即为根结点 - int rootVal = postorder[postRight - 1]; - TreeNode root = new TreeNode(rootVal); - int rootIndex = 0; - // 根据根结点的值找到该值在中序数组inorder里的位置 - for (int i = inLeft; i < inRight; i++) { - if (inorder[i] == rootVal) { - rootIndex = i; - break; - } - } - // 根据rootIndex划分左右子树 - root.left = buildTree1(inorder, inLeft, rootIndex, - postorder, postLeft, postLeft + (rootIndex - inLeft)); - root.right = buildTree1(inorder, rootIndex + 1, inRight, - postorder, postLeft + (rootIndex - inLeft), postRight - 1); + int rootIndex = map.get(postorder[postEnd - 1]); // 找到后序遍历的最后一个元素在中序遍历中的位置 + TreeNode root = new TreeNode(inorder[rootIndex]); // 构造结点 + int lenOfLeft = rootIndex - inBegin; // 保存中序左子树个数，用来确定后序数列的个数 + root.left = findNode(inorder, inBegin, rootIndex, + postorder, postBegin, postBegin + lenOfLeft); + root.right = findNode(inorder, rootIndex + 1, inEnd, + postorder, postBegin + lenOfLeft, postEnd - 1); + return root; } } @@ -622,31 +616,29 @@ class Solution { ```java class Solution { + Map map; public TreeNode buildTree(int[] preorder, int[] inorder) { - return helper(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1); - } - - public TreeNode helper(int[] preorder, int preLeft, int preRight, - int[] inorder, int inLeft, int inRight) { - // 递归终止条件 - if (inLeft > inRight || preLeft > preRight) return null; - - // val 为前序遍历第一个的值，也即是根节点的值 - // idx 为根据根节点的值来找中序遍历的下标 - int idx = inLeft, val = preorder[preLeft]; - TreeNode root = new TreeNode(val); - for (int i = inLeft; i <= inRight; i++) { - if (inorder[i] == val) { - idx = i; - break; - } + map = new HashMap<>(); + for (int i = 0; i < inorder.length; i++) { // 用map保存中序序列的数值对应位置 + map.put(inorder[i], i); } - // 根据 idx 来递归找左右子树 - root.left = helper(preorder, preLeft + 1, preLeft + (idx - inLeft), - inorder, inLeft, idx - 1); - root.right = helper(preorder, preLeft + (idx - inLeft) + 1, preRight, - inorder, idx + 1, inRight); + return findNode(preorder, 0, preorder.length, inorder, 0, inorder.length); // 前闭后开 + } + + public TreeNode findNode(int[] preorder, int preBegin, int preEnd, int[] inorder, int inBegin, int inEnd) { + // 参数里的范围都是前闭后开 + if (preBegin >= preEnd || inBegin >= inEnd) { // 不满足左闭右开，说明没有元素，返回空树 + return null; + } + int rootIndex = map.get(preorder[preBegin]); // 找到前序遍历的第一个元素在中序遍历中的位置 + TreeNode root = new TreeNode(inorder[rootIndex]); // 构造结点 + int lenOfLeft = rootIndex - inBegin; // 保存中序左子树个数，用来确定前序数列的个数 + root.left = findNode(preorder, preBegin + 1, preBegin + lenOfLeft + 1, + inorder, inBegin, rootIndex); + root.right = findNode(preorder, preBegin + lenOfLeft + 1, preEnd, + inorder, rootIndex + 1, inEnd); + return root; } }