說實話我還是沒看懂
收穫:
1.在計算機科學中,二叉樹是每個結點最多有兩個子樹的樹結構。通常子樹被稱作“左子樹”(left subtree)和“右子樹”(right subtree)
2.一棵深度爲k,且有2^k-1個結點的二叉樹,稱爲滿二叉樹,這種樹的特點是每一層上的結點數都是最大結點數3.在一棵二叉樹中,除最後一層外,若其餘層都是滿的,並且或者最後一層是滿的,或者是在右邊缺少連續若干結點,則此二叉樹爲完全二叉樹.如下圖,若6缺失則爲不完全二叉樹,若7缺失則仍爲完全二叉樹
以下圖爲例:
- 前序遍歷:根 左 右 1234567
- 中序遍歷:左 根 右 4251367
- 層次遍歷:按層級由左到右遍歷 1234567
![在這裏插入圖片描述](https://img-blog.csdnimg.cn/20200326132915432.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3kxODc5MTA1MDc3OQ==,size_16,color_FFFFFF,t_70
重建二叉樹:
輸入某二叉樹的前序遍歷和中序遍歷的結果,請重建該二叉樹。假設輸入的前序遍歷和中序遍歷的結果中都不含重複的數字。
給出
前序遍歷 :3,9,20,15,7
中序遍歷 :9,3,15,20,7
遞歸實現:
class Solution {
public TreeNode buildTree(int[] preorder, int[] inorder) {
if (preorder == null || preorder.length == 0) {
return null;
}
Map<Integer, Integer> indexMap = new HashMap<Integer, Integer>();
int length = preorder.length;
for (int i = 0; i < length; i++) {
indexMap.put(inorder[i], i);
}
TreeNode root = buildTree(preorder, 0, length - 1, inorder, 0, length - 1, indexMap);
return root;
}
public TreeNode buildTree(int[] preorder, int preorderStart, int preorderEnd, int[] inorder, int inorderStart, int inorderEnd, Map<Integer, Integer> indexMap) {
if (preorderStart > preorderEnd) {
return null;
}
int rootVal = preorder[preorderStart];
TreeNode root = new TreeNode(rootVal);
if (preorderStart == preorderEnd) {
return root;
} else {
int rootIndex = indexMap.get(rootVal);
int leftNodes = rootIndex - inorderStart, rightNodes = inorderEnd - rootIndex;
TreeNode leftSubtree = buildTree(preorder, preorderStart + 1, preorderStart + leftNodes, inorder, inorderStart, rootIndex - 1, indexMap);
TreeNode rightSubtree = buildTree(preorder, preorderEnd - rightNodes + 1, preorderEnd, inorder, rootIndex + 1, inorderEnd, indexMap);
root.left = leftSubtree;
root.right = rightSubtree;
return root;
}
}
}
迭代實現
class Solution {
public TreeNode buildTree(int[] preorder, int[] inorder) {
if (preorder == null || preorder.length == 0) {
return null;
}
TreeNode root = new TreeNode(preorder[0]);
int length = preorder.length;
Stack<TreeNode> stack = new Stack<TreeNode>();
stack.push(root);
int inorderIndex = 0;
for (int i = 1; i < length; i++) {
int preorderVal = preorder[i];
TreeNode node = stack.peek();
if (node.val != inorder[inorderIndex]) {
node.left = new TreeNode(preorderVal);
stack.push(node.left);
} else {
while (!stack.isEmpty() && stack.peek().val == inorder[inorderIndex]) {
node = stack.pop();
inorderIndex++;
}
node.right = new TreeNode(preorderVal);
stack.push(node.right);
}
}
return root;
}
}