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修改0106从中序与后序遍历序列构造二叉树 Java版本
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
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@ -584,35 +584,29 @@ tree2 的前序遍历是[1 2 3], 后序遍历是[3 2 1]。
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```java
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```java
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class Solution {
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class Solution {
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Map<Integer, Integer> map; // 方便根据数值查找位置
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public TreeNode buildTree(int[] inorder, int[] postorder) {
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public TreeNode buildTree(int[] inorder, int[] postorder) {
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return buildTree1(inorder, 0, inorder.length, postorder, 0, postorder.length);
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map = new HashMap<>();
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for (int i = 0; i < inorder.length; i++) { // 用map保存中序序列的数值对应位置
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map.put(inorder[i], i);
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}
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return findNode(inorder, 0, inorder.length, postorder,0, postorder.length); // 前闭后开
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}
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}
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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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public TreeNode findNode(int[] inorder, int inBegin, int inEnd, int[] postorder, int postBegin, int postEnd) {
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// 没有元素了
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// 参数里的范围都是前闭后开
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if (inRight - inLeft < 1) {
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if (inBegin >= inEnd || postBegin >= postEnd) { // 不满足左闭右开,说明没有元素,返回空树
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return null;
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return null;
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}
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}
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// 只有一个元素了
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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
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if (inRight - inLeft == 1) {
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TreeNode root = new TreeNode(inorder[rootIndex]); // 构造结点
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return new TreeNode(inorder[inLeft]);
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int lenOfLeft = rootIndex - inBegin; // 保存中序左子树个数,用来确定后序数列的个数
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}
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root.left = findNode(inorder, inBegin, rootIndex,
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// 后序数组postorder里最后一个即为根结点
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postorder, postBegin, postBegin + lenOfLeft);
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int rootVal = postorder[postRight - 1];
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root.right = findNode(inorder, rootIndex + 1, inEnd,
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TreeNode root = new TreeNode(rootVal);
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postorder, postBegin + lenOfLeft, postEnd - 1);
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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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break;
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}
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}
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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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return root;
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}
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}
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}
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}
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@ -622,31 +616,29 @@ class Solution {
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```java
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```java
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class Solution {
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class Solution {
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Map<Integer, Integer> map;
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public TreeNode buildTree(int[] preorder, int[] inorder) {
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public TreeNode buildTree(int[] preorder, int[] inorder) {
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return helper(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1);
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map = new HashMap<>();
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}
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for (int i = 0; i < inorder.length; i++) { // 用map保存中序序列的数值对应位置
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map.put(inorder[i], i);
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public TreeNode helper(int[] preorder, int preLeft, int preRight,
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int[] inorder, int inLeft, int inRight) {
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// 递归终止条件
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if (inLeft > inRight || preLeft > preRight) return null;
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// val 为前序遍历第一个的值,也即是根节点的值
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// idx 为根据根节点的值来找中序遍历的下标
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int idx = inLeft, val = preorder[preLeft];
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TreeNode root = new TreeNode(val);
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for (int i = inLeft; i <= inRight; i++) {
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if (inorder[i] == val) {
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idx = i;
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break;
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}
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}
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}
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// 根据 idx 来递归找左右子树
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return findNode(preorder, 0, preorder.length, inorder, 0, inorder.length); // 前闭后开
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root.left = helper(preorder, preLeft + 1, preLeft + (idx - inLeft),
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}
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inorder, inLeft, idx - 1);
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root.right = helper(preorder, preLeft + (idx - inLeft) + 1, preRight,
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public TreeNode findNode(int[] preorder, int preBegin, int preEnd, int[] inorder, int inBegin, int inEnd) {
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inorder, idx + 1, inRight);
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// 参数里的范围都是前闭后开
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if (preBegin >= preEnd || inBegin >= inEnd) { // 不满足左闭右开,说明没有元素,返回空树
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return null;
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}
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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
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TreeNode root = new TreeNode(inorder[rootIndex]); // 构造结点
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int lenOfLeft = rootIndex - inBegin; // 保存中序左子树个数,用来确定前序数列的个数
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root.left = findNode(preorder, preBegin + 1, preBegin + lenOfLeft + 1,
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inorder, inBegin, rootIndex);
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root.right = findNode(preorder, preBegin + lenOfLeft + 1, preEnd,
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inorder, rootIndex + 1, inEnd);
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return root;
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return root;
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}
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}
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}
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}
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