1. 最大二叉树（中等）
地址: https://leetcode-cn.com/problems/maximum-binary-tree/
2021/11/30
做题反思：length
class Solution {
public TreeNode constructMaximumBinaryTree(int[] nums) {
return build(nums, 0, nums.length - 1);
}
TreeNode build(int[] nums, int lo, int hi) {
if (lo > hi) {
return null;
}
int index = -1, max = Integer.MIN_VALUE;
for (int i = lo; i <= hi; i++) {
if (nums[i] > max) {
max = nums[i];
index = i;
}
}
TreeNode root = new TreeNode(max);
root.left = build(nums, lo, index - 1);
root.right = build(nums, index + 1, hi);
return root;
}
}
2. 从前序与中序遍历序列构造二叉树（中等）
地址: https://leetcode-cn.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/
2021/11/30
做题反思：
class Solution {
public TreeNode buildTree(int[] preorder, int[] inorder) {
return build(preorder, 0, preorder.length - 1,
inorder, 0, inorder.length - 1);
}
TreeNode build(int[] preorder, int preStart, int preEnd,
int[] inorder, int inStart, int inEnd) {
if (preStart > preEnd || inStart > inEnd) {
return null;
}
int rootVal = preorder[preStart];
int index = 0;
for (int i = inStart; i <= inEnd; i++) {
if (inorder[i] == rootVal) {
index = i;
break;
}
}
TreeNode root = new TreeNode(rootVal);
int leftSize = index - inStart;
root.left = build(preorder, preStart + 1, preStart + leftSize,
inorder, inStart, index - 1);
root.right = build(preorder, preStart + leftSize + 1, preEnd,
inorder, index + 1, inEnd);
return root;
}
3. 从中序与后序遍历序列构造二叉树 （中等）
地址: https://leetcode-cn.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/
2021/11/30
做题反思：
class Solution {
public TreeNode buildTree(int[] inorder, int[] postorder) {
return build(inorder, 0, inorder.length - 1,
postorder, 0, postorder.length - 1);
}
TreeNode build(int[] inorder, int inStart, int inEnd,
int[] postorder, int postStart, int postEnd) {
if (inStart > inEnd || postStart > postEnd) {
return null;
}
int rootVal = postorder[postEnd];
int index = 0;
for (int i = inStart; i <= inEnd; i++) {
if (inorder[i] == rootVal) {
index = i;
break;
}
}
TreeNode root = new TreeNode(rootVal);
int leftSize = index - inStart;
root.left = build(inorder, inStart, index - 1,
postorder, postStart, postStart + leftSize - 1);
root.right = build(inorder, index + 1, inEnd,
postorder, postStart + leftSize, postEnd - 1);
return root;
}
}
