Merge pull request #812 from casnz1601/patch-1

Update 0110.平衡二叉树.md
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程序员Carl
2021-10-05 14:22:23 +08:00
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@ -125,9 +125,10 @@ public:
1. 明确递归函数的参数和返回值
参数的话为传入节点指针,就没有其他参数需要传递了,返回值要返回传入节点为根节点树的深度
参数:当前传入节点。
返回值:以当前传入节点为根节点的树的高度。
那么如何标记左右子树是否差值大于1呢
那么如何标记左右子树是否差值大于1呢
如果当前传入节点为根节点的二叉树已经不是二叉平衡树了,还返回高度的话就没有意义了。
@ -136,9 +137,9 @@ public:
代码如下:
```
```CPP
// -1 表示已经不是平衡二叉树了,否则返回值是以该节点为根节点树的高度
int getDepth(TreeNode* node)
int getHeight(TreeNode* node)
```
2. 明确终止条件
@ -147,7 +148,7 @@ int getDepth(TreeNode* node)
代码如下:
```
```CPP
if (node == NULL) {
return 0;
}
@ -155,23 +156,23 @@ if (node == NULL) {
3. 明确单层递归的逻辑
如何判断当前传入节点为根节点的二叉树是否是平衡二叉树呢当然是左子树高度和右子树高度相差
如何判断当前传入节点为根节点的二叉树是否是平衡二叉树呢当然是左子树高度和右子树高度的差值
分别求出左右子树的高度然后如果差值小于等于1则返回当前二叉树的高度否则则返回-1表示已经不是二叉树了。
分别求出左右子树的高度然后如果差值小于等于1则返回当前二叉树的高度否则则返回-1表示已经不是二叉平衡树了。
代码如下:
```CPP
int leftDepth = depth(node->left); // 左
if (leftDepth == -1) return -1;
int rightDepth = depth(node->right); // 右
if (rightDepth == -1) return -1;
int leftHeight = getHeight(node->left); // 左
if (leftHeight == -1) return -1;
int rightHeight = getHeight(node->right); // 右
if (rightHeight == -1) return -1;
int result;
if (abs(leftDepth - rightDepth) > 1) { // 中
if (abs(leftHeight - rightHeight) > 1) { // 中
result = -1;
} else {
result = 1 + max(leftDepth, rightDepth); // 以当前节点为根节点的最大高度
result = 1 + max(leftHeight, rightHeight); // 以当前节点为根节点的树的最大高度
}
return result;
@ -180,27 +181,27 @@ return result;
代码精简之后如下:
```CPP
int leftDepth = getDepth(node->left);
if (leftDepth == -1) return -1;
int rightDepth = getDepth(node->right);
if (rightDepth == -1) return -1;
return abs(leftDepth - rightDepth) > 1 ? -1 : 1 + max(leftDepth, rightDepth);
int leftHeight = getHeight(node->left);
if (leftHeight == -1) return -1;
int rightHeight = getHeight(node->right);
if (rightHeight == -1) return -1;
return abs(leftHeight - rightHeight) > 1 ? -1 : 1 + max(leftHeight, rightHeight);
```
此时递归的函数就已经写出来了,这个递归的函数传入节点指针,返回以该节点为根节点的二叉树的高度,如果不是二叉平衡树,则返回-1。
getDepth整体代码如下:
getHeight整体代码如下:
```CPP
int getDepth(TreeNode* node) {
int getHeight(TreeNode* node) {
if (node == NULL) {
return 0;
}
int leftDepth = getDepth(node->left);
if (leftDepth == -1) return -1;
int rightDepth = getDepth(node->right);
if (rightDepth == -1) return -1;
return abs(leftDepth - rightDepth) > 1 ? -1 : 1 + max(leftDepth, rightDepth);
int leftHeight = getHeight(node->left);
if (leftHeight == -1) return -1;
int rightHeight = getHeight(node->right);
if (rightHeight == -1) return -1;
return abs(leftHeight - rightHeight) > 1 ? -1 : 1 + max(leftHeight, rightHeight);
}
```
@ -210,18 +211,18 @@ int getDepth(TreeNode* node) {
class Solution {
public:
// 返回以该节点为根节点的二叉树的高度,如果不是二叉搜索树了则返回-1
int getDepth(TreeNode* node) {
int getHeight(TreeNode* node) {
if (node == NULL) {
return 0;
}
int leftDepth = getDepth(node->left);
if (leftDepth == -1) return -1; // 说明左子树已经不是二叉平衡树
int rightDepth = getDepth(node->right);
if (rightDepth == -1) return -1; // 说明右子树已经不是二叉平衡树
return abs(leftDepth - rightDepth) > 1 ? -1 : 1 + max(leftDepth, rightDepth);
int leftHeight = getHeight(node->left);
if (leftHeight == -1) return -1;
int rightHeight = getHeight(node->right);
if (rightHeight == -1) return -1;
return abs(leftHeight - rightHeight) > 1 ? -1 : 1 + max(leftHeight, rightHeight);
}
bool isBalanced(TreeNode* root) {
return getDepth(root) == -1 ? false : true;
return getHeight(root) == -1 ? false : true;
}
};
```
@ -498,20 +499,35 @@ class Solution {
## Python
递归法:
```python
```python3
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def isBalanced(self, root: TreeNode) -> bool:
return True if self.getDepth(root) != -1 else False
if self.get_height(root) != -1:
return True
else:
return False
#返回以该节点为根节点的二叉树的高度,如果不是二叉搜索树了则返回-1
def getDepth(self, node):
if not node:
def get_height(self, root: TreeNode) -> int:
# Base Case
if not root:
return 0
leftDepth = self.getDepth(node.left)
if leftDepth == -1: return -1 #说明左子树已经不是二叉平衡树
rightDepth = self.getDepth(node.right)
if rightDepth == -1: return -1 #说明右子树已经不是二叉平衡树
return -1 if abs(leftDepth - rightDepth)>1 else 1 + max(leftDepth, rightDepth)
# 左
if (left_height := self.get_height(root.left)) == -1:
return -1
# 右
if (right_height := self.get_height(root.right)) == -1:
return -1
# 中
if abs(left_height - right_height) > 1:
return -1
else:
return 1 + max(left_height, right_height)
```
迭代法: