算法：树Z型遍历

最新推荐文章于 2025-04-03 11:02:57 发布

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这段代码实现了一个二叉树的层次遍历，并以特定顺序打印节点。首先从根节点开始，然后按照右子节点优先的原则遍历每一层。在遍历过程中，先将当前层级的节点加入队列，然后反向处理并打印，最后添加下一层节点到队列中。

__doc__ = u"""
            1
      2        3
    4    5  6     7
   8  9         10  11 
                
"""
import copy

class Node(object):
    def __init__(self,value):
        self.value = value
        self.left = None
        self.right = None
    
    def __str__(self):
        return str(self.value)
        
        
def print_bin_tree(root):
    array = []
    
    tmp = [{"node":root,"level":0}]
    
    
    while tmp:
        array = copy.copy(tmp)
        tmp = []
        
        for elem in array: #打印顺序
            node = elem["node"]
            level = elem["level"]
            print(node.value)
        
        array.reverse()
        
        for elem  in array: #添加顺序
            node = elem["node"]
            level = elem["level"]
            
            if level %2 == 0: #先右，后左
                if node.right:
                    tmp.append({"node":node.right,"level":level + 1})
                
                if node.left:
                    tmp.append({"node":node.left,"level":level+1})
            else:
                if node.left:
                    tmp.append({"node":node.left,"level":level+1})
                
                if node.right:
                    tmp.append({"node":node.right,"level":level+1})
            
        level = level + 1
        

if __name__ == "__main__":
    root = Node(1)
    root.left = Node(2)
    root.right = Node(3)
    root.left.left = Node(4)
    root.left.right = Node(5)
    root.right.left = Node(6)
    root.right.right = Node(7)
    root.left.left.left = Node(8)
    root.left.left.right = Node(9)
    
    root.right.right.left = Node(10)
    root.right.right.right = Node(11)
    
    print_bin_tree(root)