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Merge pull request #95 from Joshua-Lu/patch-11
更新 0106.从中序与后序遍历序列构造二叉树 Java版本
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@ -585,46 +585,38 @@ Java:
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```java
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class Solution {
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public TreeNode buildTree(int[] inorder, int[] postorder) {
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if (inorder.length == 0 || postorder.length == 0) return null;
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return buildTree(inorder, postorder,0,inorder.length,0,postorder.length);
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return buildTree1(inorder, 0, inorder.length, postorder, 0, postorder.length);
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}
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private TreeNode buildTree(int[] inorder, int[] postorder, int infrom, int into,
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int postfrom, int postto) {
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if (postfrom == postto) return null;
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int rootValue = postorder[postto - 1];
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TreeNode root = new TreeNode(rootValue);
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if (postfrom + 1 == postto) return root;
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int splitNum = postto - 1;
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for (int i = infrom; i < into; i++) {
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if (inorder[i] == rootValue) {
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splitNum = i;
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break;
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public TreeNode buildTree1(int[] inorder, int inLeft, int inRight,
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int[] postorder, int postLeft, int postRight) {
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// 没有元素了
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if (inRight - inLeft < 1) {
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return null;
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}
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// 只有一个元素了
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if (inRight - inLeft == 1) {
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return new TreeNode(inorder[inLeft]);
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}
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// 后序数组postorder里最后一个即为根结点
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int rootVal = postorder[postRight - 1];
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TreeNode root = new TreeNode(rootVal);
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int rootIndex = 0;
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// 根据根结点的值找到该值在中序数组inorder里的位置
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for (int i = inLeft; i < inRight; i++) {
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if (inorder[i] == rootVal) {
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rootIndex = i;
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}
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}
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int inLeftBegin = infrom;
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int inLeftEnd = splitNum;
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int inRightBegin = splitNum + 1;
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int inRightEnd = into;
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int postLeftBegin = postfrom;
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int postLeftEnd = postLeftBegin + (splitNum - inLeftBegin);
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int postRightBegin = postLeftBegin + (splitNum - inLeftBegin);
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int postRightEnd = postto - 1;
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root.left = buildTree(inorder,postorder,inLeftBegin,inLeftEnd,postLeftBegin,postLeftEnd);
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root.right = buildTree(inorder,postorder,inRightBegin,inRightEnd,postRightBegin,postRightEnd);
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// 根据rootIndex划分左右子树
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root.left = buildTree1(inorder, inLeft, rootIndex,
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postorder, postLeft, postLeft + (rootIndex - inLeft));
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root.right = buildTree1(inorder, rootIndex + 1, inRight,
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postorder, postLeft + (rootIndex - inLeft), postRight - 1);
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return root;
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}
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}
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```
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Python:
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