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Question:
Given a complete binary tree, count the number of nodes.
Definition of a complete binary tree from Wikipedia:
In a complete binary tree every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible. It can have between 1 and 2h nodes inclusive at the last level h.
Analysis:
Steps to solve this problem:
1) get the height of left-most part
2) get the height of right-most part
3) when they are equal, the # of nodes = 2^h -1
4) when they are not equal, recursively get # of nodes from left&right sub-trees
public int countNodes(TreeNode root) { // Notice the traditional traversal will exceed the time limit and we also can‘t use Math.pow() cause it is too slow. if(root == null) return 0; TreeNode ln = root, rn = root; int l = 0, r = 0; while(ln != null) { l++; ln = ln.left; } while(rn != null) { r++; rn = rn.right; } // Math.pow(2,n) = 2<<(n-1) e.g. 2^1 == 2 == 2<<0 if(l == r) return (2<<(l - 1)) - 1; // notice "<<" is right connect so we need "() out of 2<<(l-1)" return 1 + countNodes(root.left) + countNodes(root.right); }
8.8 LeetCode 222 Count Complete Tree Nodes
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原文地址：http://www.cnblogs.com/michael-du/p/4714105.html
