The diameter of a binary tree

Definition:Thediameterof a tree (sometimes called the width) is the number of nodes on the longest path between two leaves in the tree.

The diameter of a treeTis the largest of the following quantities:

• the diameter ofT's left subtree
• the diameter ofT's right subtree
• the longest path between leaves that goes through the root ofT(this can be computed from the heights of the subtrees ofT)

/* Function to get diameter of a binary tree */ static int diameter(Node root) { if (root == null) { return 0; } int lHeight = height(root.leftChild); int rHeight = height(root.rightChild); int lDiameter = diameter(root.leftChild); int rDiameter = diameter(root.rightChild); return max(lHeight + rHeight + 1, max(lDiameter, rDiameter)); } /* The function Compute the "height" of a tree. Height is the number f nodes along the longest path from the root node down to the farthest leaf node.*/ static int height(Node root) { if (root == null) { return 0; } return 1 + max(height(root.leftChild), height(root.rightChild)); }

The complexity of the program is O(n^2), because the height function of called everytime when we go through the tree.

There is another solution and the complexity is O(N). The main idea of this approach is that the node stores its left child's and right child's maximum diameter if thenode's child is the "root", therefore, there is no need to recursively call the height method. The drawback is we need to add two extra variables in the node class.

int findMaxLen(Node root) { int nMaxLen = 0; if (root == null) return 0; if (root.leftChild == null) { root.nMaxLeft = 0; } if (root.rightChild == null) { root.nMaxRight = 0; } if (root.leftChild != null) { findMaxLen(root.leftChild); } if (root.rightChild != null) { findMaxLen(root.rightChild); } if (root.leftChild != null) { int nTempMaxLen = 0; nTempMaxLen = (root.leftChild.nMaxLeft > root.leftChild.nMaxRight) ? root.leftChild.nMaxLeft : root.leftChild.nMaxRight; root.nMaxLeft = nTempMaxLen + 1; } if (root.rightChild != null) { int nTempMaxLen = 0; nTempMaxLen = (root.rightChild.nMaxLeft > root.rightChild.nMaxRight) ? root.rightChild.nMaxLeft : root.rightChild.nMaxRight; root.nMaxRight = nTempMaxLen + 1; } if (root.nMaxLeft + root.nMaxRight > nMaxLen) { nMaxLen = root.nMaxLeft + root.nMaxRight; } return nMaxLen; }参考：编程之美