仅考虑偶数级叶的二叉树高度

假设只有偶数层的节点被视为有效叶节点，求二叉树的高度。
二叉树的高度是树根和最远叶子之间的边数。但是，如果我们改变一下叶子节点的定义会怎样。让我们将有效叶节点定义为没有子节点且处于偶数级别的节点（将根节点视为奇数级别节点）。

输出：树的高度为 4

解决方案：解决这个问题的方法与正常的高度查找方法略有不同。在返回步骤中，我们检查节点是否是有效的根节点。如果它有效，则返回 1，否则我们返回 0。现在在递归步骤中 - 如果左子树和右子树都产生 0，则当前节点也产生 0，因为在这种情况下，没有来自当前节点的路径到一个有效的叶节点。但是，如果孩子返回的值中至少有一个非零，则意味着该路径上的叶节点是有效的叶节点，因此该路径可以对最终结果做出贡献，因此我们返回值的最大值为当前节点返回 + 1。

C++

/* Program to find height of the tree considering
   only even level leaves. */
#include 
using namespace std;
 
/* A binary tree node has data, pointer to
   left child and a pointer to right child */
struct Node {
    int data;
    struct Node* left;
    struct Node* right;
};
 
int heightOfTreeUtil(Node* root, bool isEven)
{
    // Base Case
    if (!root)
        return 0;
 
    if (!root->left && !root->right) {
        if (isEven)
            return 1;
        else
            return 0;
    }
 
    /*left stores the result of left subtree,
      and right stores the result of right subtree*/
    int left = heightOfTreeUtil(root->left, !isEven);
    int right = heightOfTreeUtil(root->right, !isEven);
 
    /*If both left and right returns 0, it means
      there is no valid path till leaf node*/
    if (left == 0 && right == 0)
        return 0;
 
    return (1 + max(left, right));
}
 
/* Helper function that allocates a new node with the
   given data and NULL left and right pointers. */
struct Node* newNode(int data)
{
    struct Node* node =
              (struct Node*)malloc(sizeof(struct Node));
    node->data = data;
    node->left = NULL;
    node->right = NULL;
 
    return (node);
}
 
int heightOfTree(Node* root)
{
    return heightOfTreeUtil(root, false);
}
 
/* Driver program to test above functions*/
int main()
{
    // Let us create binary tree shown in above diagram
    struct Node* root = newNode(1);
 
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->left->right->left = newNode(6);
    cout << "Height of tree is " << heightOfTree(root);
    return 0;
}

Java

/* Java Program to find height of the tree considering
only even level leaves. */
class GfG {
 
/* A binary tree node has data, pointer to
left child and a pointer to right child */
static class Node {
    int data;
    Node left;
 Node right;
}
 
static int heightOfTreeUtil(Node root, boolean isEven)
{
    // Base Case
    if (root == null)
        return 0;
 
    if (root.left == null && root.right == null) {
        if (isEven == true)
            return 1;
        else
            return 0;
    }
 
    /*left stores the result of left subtree,
    and right stores the result of right subtree*/
    int left = heightOfTreeUtil(root.left, !isEven);
    int right = heightOfTreeUtil(root.right, !isEven);
 
    /*If both left and right returns 0, it means
    there is no valid path till leaf node*/
    if (left == 0 && right == 0)
        return 0;
 
    return (1 + Math.max(left, right));
}
 
/* Helper function that allocates a new node with the
given data and NULL left and right pointers. */
static Node newNode(int data)
{
    Node node = new Node();
    node.data = data;
    node.left = null;
    node.right = null;
 
    return (node);
}
 
static int heightOfTree(Node root)
{
    return heightOfTreeUtil(root, false);
}
 
/* Driver program to test above functions*/
public static void main(String[] args)
{
    // Let us create binary tree shown in above diagram
    Node root = newNode(1);
 
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.left.right = newNode(5);
    root.left.right.left = newNode(6);
    System.out.println("Height of tree is " + heightOfTree(root));
}
}

Python3

# Program to find height of the tree considering
# only even level leaves.
 
# Helper class that allocates a new node with the
# given data and None left and right pointers.
class newNode:
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None
 
def heightOfTreeUtil(root, isEven):
     
    # Base Case
    if (not root):
        return 0
 
    if (not root.left and not root.right):
        if (isEven):
            return 1
        else:
            return 0
 
    # left stores the result of left subtree,
    # and right stores the result of right subtree
    left = heightOfTreeUtil(root.left, not isEven)
    right = heightOfTreeUtil(root.right, not isEven)
 
    #If both left and right returns 0, it means
    # there is no valid path till leaf node
    if (left == 0 and right == 0):
        return 0
 
    return (1 + max(left, right))
 
def heightOfTree(root):
    return heightOfTreeUtil(root, False)
 
# Driver Code
if __name__ == '__main__':
     
    # Let us create binary tree shown
    # in above diagram
    root = newNode(1)
 
    root.left = newNode(2)
    root.right = newNode(3)
    root.left.left = newNode(4)
    root.left.right = newNode(5)
    root.left.right.left = newNode(6)
    print("Height of tree is",
           heightOfTree(root))
 
# This code is contributed by PranchalK

C#

/* C# Program to find height of the tree considering
only even level leaves. */
using System;
 
class GfG
{
 
    /* A binary tree node has data, pointer to
    left child and a pointer to right child */
    class Node
    {
        public int data;
        public Node left;
        public Node right;
    }
 
    static int heightOfTreeUtil(Node root,
                               bool isEven)
    {
        // Base Case
        if (root == null)
            return 0;
 
        if (root.left == null &&
            root.right == null)
        {
            if (isEven == true)
                return 1;
            else
                return 0;
        }
 
        /*left stores the result of left subtree,
        and right stores the result of right subtree*/
        int left = heightOfTreeUtil(root.left, !isEven);
        int right = heightOfTreeUtil(root.right, !isEven);
 
        /*If both left and right returns 0, it means
        there is no valid path till leaf node*/
        if (left == 0 && right == 0)
            return 0;
 
        return (1 + Math.Max(left, right));
    }
 
    /* Helper function that allocates a new node with the
    given data and NULL left and right pointers. */
    static Node newNode(int data)
    {
        Node node = new Node();
        node.data = data;
        node.left = null;
        node.right = null;
 
        return (node);
    }
 
    static int heightOfTree(Node root)
    {
        return heightOfTreeUtil(root, false);
    }
 
    /* Driver code*/
    public static void Main(String[] args)
    {
        // Let us create binary tree
        // shown in above diagram
        Node root = newNode(1);
 
        root.left = newNode(2);
        root.right = newNode(3);
        root.left.left = newNode(4);
        root.left.right = newNode(5);
        root.left.right.left = newNode(6);
        Console.WriteLine("Height of tree is " +
                            heightOfTree(root));
    }
}
 
/* This code is contributed by Rajput-Ji*/

Javascript

/* javascript Program to find height of the tree considering
only even level leaves. */
 
    /*
     * A binary tree node has data, pointer to left child and a pointer to right
     * child
     */
      
     class Node {
            constructor(val) {
                this.data = val;
                this.left = null;
                this.right = null;
            }
        }
 
    function heightOfTreeUtil(root,  isEven) {
        // Base Case
        if (root == null)
            return 0;
 
        if (root.left == null && root.right == null) {
            if (isEven == true)
                return 1;
            else
                return 0;
        }
 
        /*
         * left stores the result of left subtree, and right stores the result of right
         * subtree
         */
        var left = heightOfTreeUtil(root.left, !isEven);
        var right = heightOfTreeUtil(root.right, !isEven);
 
        /*
         * If both left and right returns 0, it means there is no valid path till leaf
         * node
         */
        if (left == 0 && right == 0)
            return 0;
 
        return (1 + Math.max(left, right));
    }
 
    /*
     * Helper function that allocates a new node with the given data and NULL left
     * and right pointers.
     */
    function newNode(data) {
var node = new Node();
        node.data = data;
        node.left = null;
        node.right = null;
 
        return (node);
    }
 
    function heightOfTree(root) {
        return heightOfTreeUtil(root, false);
    }
 
    /* Driver program to test above functions */
     
        // Let us create binary tree shown in above diagram
var root = newNode(1);
 
        root.left = newNode(2);
        root.right = newNode(3);
        root.left.left = newNode(4);
        root.left.right = newNode(5);
        root.left.right.left = newNode(6);
        document.write("Height of tree is " + heightOfTree(root));
 
// This code is contributed by umadevi9616

输出：

Height of tree is 4

时间复杂度： O(n)，其中 n 是给定二叉树中的节点数。

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